§ 1. Homotopies, homeotopies

In this paragraph and those that follow, $\mathbf{I}$ denotes the interval $[0,1]$ of $\mathbf{R}$ .

1. Homotopic continuous mappings

Definition 1 01V4

Let X and Y be topological spaces and let $f$ and $g$ be continuous mappings of X into Y. A homotopy relating $f$ to $g$ is by definition a continuous mapping $\sigma : X\times \mathbf{I}\rightarrow Y$ such that, for every $x\in X$ , one has $\sigma (x,0) =f(x)$ and $\sigma (x,1) =g(x)$ . One says that $f$ is homotopic to $g$ if there exists a homotopy relating $f$ to $g$ .

One says that $f$ is the origin and $g$ the term of the homotopy $\sigma ($ cf. below, III, p. 257, remark 2).

Let A be a subset of X. One says that the homotopy $\sigma$ is fixed on A if, for every $a\in A$ , the mapping $t\mapsto \sigma (a, t)$ of $\mathbf{I}$ into Y is constant. In this case, the origin and the term of $\sigma$ coincide at every point of A.

Let X and Y be topological spaces. One says that homotopies $\sigma : X\times \mathbf{I}\rightarrow Y$ and $\tau : X\times \mathbf{I}\rightarrow Y$ are juxtaposed if the term of $\sigma$ is

N. Bourbaki and Springer-Verlag Berlin Heidelberg 2 016

N.© Bourbaki, Algebraic Topology, DOI 10.1007/978-3-662-49361-8_3 229 the origin of $\tau$ , in other words if one has $\sigma (x,1) =\tau (x,0)$ for every $x\in X$ . In this case, the mapping $\sigma *\tau$ of $X\times \mathbf{I}$ into Y defined by

$\sigma (x,2t)$ for $0\leqslant t\leqslant 1/2$

(\sigma *\tau )(x, t) = \tag{1}

$\tau (x,2t-1)$ for $1/2\leqslant t\leqslant 1$

is continuous (TG, I, p. 19, prop. 4) and is a homotopy relating the origin of $\sigma$ to the term of $\tau$ . It is called the juxtaposed homotopy of the homotopies $\sigma$ and $\tau$ .

If $\sigma : X\times \mathbf{I}\rightarrow Y$ is a homotopy, the mapping $\overline{\sigma}: X\times \mathbf{I}\rightarrow Y$ defined bythe origin of $(\sigma x, t$ . One has $)\mapsto \frac{\sigma}{\sigma}(=x,1\sigma -$ . If $t)\sigma$ is a homotopy relating the term ofand $\tau$ are juxtaposed homotopies $\sigma$ to

of $X\times \mathbf{I}$ into Y, the homotopies $\overline{\tau}$ and $\overline{\sigma}$ are juxtaposed and one has $\overline{\sigma*\tau}=\overline{\tau}*\overline{\sigma}$ .

Proposition 1 01V5

Let X and Y be topological spaces. The relation “ $f$ is homotopic to $g$ ” is an equivalence relation in the set $\mathscr{C}(X; Y)$ of continuous mappings of X into Y.

Let $f$ be an element of $\mathscr{C}(X; Y)$ . The mapping $f\circ$ pr $_1: X\times \mathbf{I}\rightarrow Y$ is a homotopy relating $f$ to $f$ ; this relation is therefore reflexive.

Let $f$ and $g$ be elements of $\mathscr{C}(X; Y)$ and $\sigma : X\times \mathbf{I}\rightarrow Y$ a homotopy joining $f$ to $g$ . The mapping $\overline{\sigma}: X\times \mathbf{I}\rightarrow Y$ is then a homotopy joining $g$ to $f$ ; the relation considered is therefore symmetric.

Finally, let us prove that it is transitive. If $f,g$ and $h$ are elements of $\mathscr{C}(X; Y),\sigma$ a homotopy joining $f$ to $g$ and $\tau$ a homotopy joining $g$ to $h$ , then $\sigma$ and $\tau$ are juxtaposed and $\sigma *\tau$ is a homotopy joining $f$ to $h$ .

Let X and Y be topological spaces. The equivalence relation “ $f$ is homotopic to $g$ ” in $\mathscr{C}(X; Y)$ (Prop. 1) is called the homotopy relation. The quotient set of $\mathscr{C}(X; Y)$ by this relation is denoted by [X; Y]. Its elements are called homotopy classes of continuous mappings of X into Y. The homotopy class of a continuous mapping $f: X\rightarrow Y$ will often be denoted by $[f]$ .

Proposition 2 01V6

Let X, Y and Z be topological spaces, $f$ and $f'$ continuous mappings of X into Y $,g$ and $g'$ continuous mappings of Y into Z. If $f$ is homotopic to $f'$ and if $g$ is homotopic to $g'$ , then $g\circ f$ is homotopic to $g'\circ f'$ .

Let $\sigma$ be a homotopy joining $f$ to $f'$ and $\tau$ a homotopy joining $g$ to $g'$ . Then the mapping $\theta : X\times \mathbf{I}\rightarrow Z$ defined by $\theta (x, t) =\tau (\sigma (x, t), t)$ is a homotopy joining $g\circ f$ to $g'\circ f'$ .

Let X, Y and Z be topological spaces. Given homotopy classes $\varphi \in [X; Y]$ and $\psi \in [Y; Z]$ , the mappings $g\circ f: X\rightarrow Z$ , where $f\in \varphi ,g\in \psi$ , all belong to one and the same homotopy class (Prop. 2), which is denoted by $\psi \circ \varphi$ and is called the composite homotopy class of the classes $\psi$ and $\varphi$ . The mapping $[X; Y]\times [Y; Z]\rightarrow [X; Z]$ which associates $\psi \circ \varphi$ to $(\varphi , \psi )$ is called the composition mapping.

Let $\varphi \in [X; Y]$ ; we have $\varphi =\varphi \circ$ [Id $_X] =$ [Id $_Y]\circ \varphi$ .

Let X, Y, Z and T be topological spaces, let $\varphi \in [X; Y]$ , $\psi \in [Y; Z]$ and $\chi \in [Z; T]$ . The homotopy class $\chi \circ (\psi \circ \varphi )$ is equal to $(\chi \circ \psi )\circ \varphi$ ; it is denoted by $\chi \circ \psi \circ \varphi$ .

Proposition 3 01V7

Let X be a topological space and let $(Y_j)_{j\in J}$ be a family of topological spaces. The mapping of $[X;\prod_{j\in J}Y_j]$ into the product set $\prod_{j\in J}[X; Y_j]$ defined by $\varphi \mapsto$ ([pr $_j]\circ \varphi$ ) $_{j\in J}$ is bijective.

Surjectivity results immediately from I, p. 25, prop. 1. Let us prove injectivity. Let $f$ and $g$ be continuous mappings of X into $\prod_{j\in J}Y_j$ . For each $j\in J$ , put $f_j=$ pr $_j\circ f$ and $g_j=$ pr $_j\circ g$ ; suppose $f_j$ homotopic to $g_j$ and let $\sigma_j$ be a homotopy joining $f_j$ to $g_j$ . The mapping $\sigma = (\sigma_j)$ of $X\times \mathbf{I}$ into $\prod_{j\in J}Y_j$ is continuous (loc. cit.) ; it is a homotopy joining $f$ to $g$ , whence the proposition.

Corollary 01V8

Let $(X_j)_{j\in J}$ and $(Y_j)_{j\in J}$ be families of topological spaces having the same index set. For each $j\in J$ , let $f_j$ and $g_j$ be continuous mappings of $X_j$ into $Y_j$ . If, for each $j\in J$ , the mappings $f_j$ and $g_j$ are homotopic, the same is true of the product mappings $f: (x_j)\mapsto (f_j(x_j))$ and $g: (x_j)\mapsto (g_j(x_j))$ of $\prod_{j\in J}X_j$ into $\prod_{j\in J}Y_j$ .

Suppose that we have $[f_j] = [g_j]$ for each $j\in J$ . We have [pr $_j]\circ [f] =$ $[f_j\circ$ pr $_j]$ and [pr $_j]\circ [g] = [g_j\circ$ pr $_j]$ , hence [pr $_j]\circ [f] =$ [pr $_j]\circ [g]$ by prop. 2, whence $[f] = [g]$ by prop. 3.

2. Pointed homotopies

Let X and Y be topological spaces and let $x$ be a point of X. A homotopy $\sigma : X\times \mathbf{I}\rightarrow Y$ is said to be pointed at $x$ if it is fixed on $\{x\}$ , that is to say if the mapping $t\mapsto \sigma (x, t)$ from $\mathbf{I}$ into Y is constant. The juxtaposition of two homotopies pointed at $x$ is pointed at $x$ . The origin and the end of a homotopy pointed at $x$ take the same value $y$ at $x$ ; they are therefore pointed continuous mappings from $(X, x)$ into $(Y, y)$ .

Let $(X, x)$ and $(Y, y)$ be pointed topological spaces (I, p. 120, definition 1). The relation « $f$ is related to $g$ by a homotopy pointed at $x$ » is an equivalence relation in the set $\mathscr{C}((X, x); (Y, y))$ , called the pointed homotopy relation. The quotient set of $\mathscr{C}((X, x); (Y, y))$ by this equivalence relation is denoted by $[(X, x); (Y, y)]$ . Its elements are called pointed homotopy classes of pointed continuous mappings from $(X, x)$ into $(Y, y)$ .

The set $\mathscr{C}((X, x); (Y, y))$ is a subset of $\mathscr{C}(X; Y)$ . It should be noted that pointed homotopy is a finer equivalence relation than the relation induced by homotopy in $\mathscr{C}(X; Y)$ . It is in general a strictly finer relation (see III, p. 321, exerc. 1 and III, p. 234, example 3).

Let also $(Z, z)$ be a pointed topological space. Let $f$ and $f'$ be pointed continuous mappings from $(X, x)$ into $(Y, y),g$ and $g'$ pointed continuous mappings from $(Y, y)$ into $(Z, z)$ . If $f$ and $f'$ are related by a homotopy $\sigma$ pointed at $x$ , and if $g$ and $g'$ are related by a homotopy $\tau$ pointed at $y$ , then $g\circ f$ and $g'\circ f'$ are related by the homotopy $\theta : X\times \mathbf{I}\rightarrow Z$ , $(u, t)\mapsto \tau (\sigma (u, t), t)$ , which is pointed at $x$ . As above, this makes it possible to define the composition mapping of $[(X, x); (Y, y)]\times [(Y, y); (Z, z)]$ into $[(X, x); (Z, z)]$ . We leave to the reader the task of formulating and proving for pointed mappings and homotopies the statements analogous to prop. 3 and its corollary.

3. Homotopic spaces

Definition 2 01V9

Let X and Y be topological spaces. A homotopy class $\varphi \in$ [X; Y] is said to be invertible if there exists a homotopy class $\psi \in [Y; X]$ such that $\psi \circ \varphi =$ [Id $_X]$ and $\varphi \circ \psi =$ [Id $_Y]$ . A continuous mapping is said to be a homotopy equivalence if its homotopy class is invertible.

Let $\varphi \in$ [X; Y] be an invertible homotopy class. There exists a unique homotopy class $\psi \in [Y; X]$ having the properties of definition 2. Let indeed $\psi ,\psi '$ be classes having these properties; one has

$\psi =\psi \circ$ [Id $_Y] =\psi \circ \varphi \circ \psi '=$ [Id $_X]\circ \psi '=\psi '$ .

This unique class is called the inverse of the homotopy class $\varphi$ and is denoted $\varphi^{-1}$ .

Let Z be a topological space. We have $\chi \circ \varphi \circ \varphi^{-1}=\chi$ for all $\chi \in$ [Y; Z] and $\theta \circ \varphi^{-1}\circ \varphi =\theta$ for all $\theta \in [X; Z]$ . It follows that the mappings $\chi \mapsto \chi \circ \varphi$ of [Y; Z] into [X; Z] and $\theta \mapsto \theta \circ \varphi^{-1}$ of [X; Z] into [Y; Z] are bijections inverse to one another. Analogously, the mapping $\chi \mapsto \varphi \circ \chi$ of [Z; X] into [Z; Y] is bijective and its inverse bijection is the mapping $\theta \mapsto \varphi^{-1}\circ \theta$ of [Z; Y] into [Z; X].

Let X, Y, Z be topological spaces, $\varphi \in [X; Y]$ and $\psi \in [Y; Z]$ invertible homotopy classes. Then the class $\psi \circ \varphi$ is invertible, with inverse $\varphi^{-1}\circ \psi^{-1}$ . Indeed, we have

(\psi \circ \varphi )\circ (\varphi^{-1}\circ \psi^{-1}) =\psi \circ (\varphi \circ \varphi^{-1})\circ \psi^{-1}

$=\psi \circ$ [Id $_Y]\circ \psi^{-1}$

$=\psi \circ \psi^{-1}=$ [Id $_Z]$

and, analogously, $(\varphi^{-1}\circ \psi^{-1})\circ (\psi \circ \varphi ) =$ [Id $_X]$ .

Let X, Y be topological spaces and let $f: X\rightarrow Y$ be a homotopy equivalence. Any continuous mapping $g: Y\rightarrow X$ whose class is the inverse of that of $f$ is called a reciprocal (or inverse) of $f$ up to homotopy. Such a mapping $g$ is a homotopy equivalence.

A homeomorphism $f$ is a homotopy equivalence, and $[f]^{-1}= [f^{-1}]$ . The composite mapping of two homotopy equivalences is a homotopy equivalence. The relation "X and Y are topological spaces and there exists a homotopy equivalence of X into Y" is an equivalence relation.

Definition 3 01VA

One says that topological spaces X and Y are homotopy equivalent if there exists a homotopy equivalence of X into Y.

Example 1 01VB

The empty topological space is homotopy equivalent only to itself.

Example 2 01VC

Let X be a nonempty topological space. In order that X be homotopy equivalent to a space reduced to one point, it is necessary and sufficient that there exist a constant mapping $p: X\rightarrow$ X which is homotopic to the identity mapping of X. (This is then so for every constant mapping $q: X\rightarrow X$ , since $[p] = [p]\circ [q] =$ [Id $_X]\circ [q] = [q]$ .) Indeed, let P be a space reduced to one point, $f: P\rightarrow X$ a mapping and $g: X\rightarrow P$ the unique mapping of X into P. We have $g\circ f=$ Id $_P$ ; in order that $f$ be a homotopy equivalence, it is necessary and sufficient that $f\circ g$ be homotopic to Id $_X$ . Now, $f\circ g$ is constant, of image $f(P)$ .

Example 3 01VD

One says that a pointed topological space $(X, x)$ is contractible, or that the topological space X is contractible at $x$ , if there exists a homotopy pointed at $x$ connecting Id $_X$ to the constant mapping of X into X of image $\{x\}$ . Such a space is homotopy equivalent to a point. There nevertheless exist spaces homotopy equivalent to a point which are not contractible at any of their points, and spaces contractible at one point, but not at every point (III, p. 321, exerc. 1).

Example 4 01VE

Let E be the numerical space of $n$ dimensions (or, more generally, a topological vector space over $\mathbf{R}$ ) and let X be a subset of E. One says that the set X is star-shaped at a point $x$ of X if, for all $y\in X$ and all $t\in \mathbf{I}$ , the point $tx+ (1-t)y$ belongs to X. A convex subset (I, p. 122) of E is star-shaped at each of its points.

A topological subspace X of E which is star-shaped at one of its points $x$ is contractible at this point. Indeed, the mapping $\sigma : X\times \mathbf{I}\rightarrow X$ defined by $\sigma (y, t) =tx+ (1-t)y$ is a pointed homotopy at $x$ joining Id $_X$ to the constant mapping with image $\{x\}$ .

In particular, every interval of $\mathbf{R}$ , every convex subset of a numerical space or, more generally, of a topological vector space over $\mathbf{R}$ is contractible at each of its points.

Example 5 01VF

We shall prove later (TA, V) that the Euclidean spheres $\mathbf{S}_n,n\geqslant 1$ , are not spaces homotopic to a point. The unit sphere of a Hilbert space of countable type and infinite dimension is contractible at each of its points (EVT, V, p. 71, exerc. 13).

4. Relative homotopies

Let X, Y be topological spaces, let A be a subspace of X and let B be a subspace of Y.

Let $f: X\rightarrow Y$ be a continuous mapping such that $f(A)\subset B$ . We say that $f$ is a homotopy of the pair $(X,A)$ onto the pair $(Y,B)$ if there exists a continuous mapping $g: Y\rightarrow X$ such that $g(B)\subset A$ , a homotopy $\sigma : X\times \mathbf{I}\rightarrow X$ , fixed on A, joining Id $_X$ to $g\circ f$ , and a homotopy $\tau : Y\times \mathbf{I}\rightarrow Y$ , fixed on B, joining Id $_Y$ to $f\circ g$ .

Suppose these conditions are satisfied. Then:

a) The mapping $f$ is a homotopy of X into Y, the mapping $g$ is a homotopy of Y into X, and these homotopies are inverses of one another up to homotopy.

b) The mapping $g$ is then a homotopy of the pair $(Y,B)$ onto the pair $(X,A)$ , called the inverse of $f$ up to homotopy.

c) The mappings $f$ and $g$ define, by passing to the subspaces, homeomorphisms of A onto B which are inverse to one another.

We say that the pairs $(X,A)$ and $(Y,B)$ are homotopic if there exists a homotopy of the pair $(X,A)$ onto the pair $(Y,B)$ . The relation “X, Y are topological spaces, A is a subspace of X, B is a subspace of Y and the pairs $(X,A)$ and $(Y,B)$ are homotopic” is an equivalence relation.

5. Retractions and contractions

Definition 4 01VG

Let X be a topological space and let A be a subset of X.

A retraction of X onto A is called a continuous mapping from X into A which is a retraction of the canonical injection of A into X. If there exists a retraction of X onto A, we say that X can be retracted onto A, or also that A is a retract of X.

Let X be a topological space and let A be a subspace of X.

Suppose that there exists a retraction $r$ of X onto A. The subspace A is identified with the set of points $x\in X$ such that $x=r(x)$ . If X is a separated space, A is then closed in X.

According to the following lemma, for a subspace A to be a retract of X, it is necessary and sufficient that every continuous mapping from A into a topological space Y extend to a continuous mapping from X into Y.

Lemma 1 01VH

Let X and Y be topological spaces and let $f: X\rightarrow Y$ be a continuous mapping. The following conditions are equivalent:

(i) For every topological space Z and every continuous mapping $g: X\rightarrow Z$ , there exists a continuous mapping $g': Y\rightarrow Z$ such that $g=$ $g'\circ f$ ;

(ii) The mapping $f$ is injective and possesses a continuous retraction ;

(iii) The mapping $f$ defines a homeomorphism of X onto its image $f(X)$ , which is a retract of Y.

Suppose that $f$ is injective and that $r: Y\rightarrow$ X is a continuous retraction of the mapping $f$ . For every continuous mapping $g: X\rightarrow Z$ , the mapping $g'=g\circ r: Y\rightarrow Z$ satisfies $g'\circ f=g\circ r\circ f=g$ . This proves that (ii) $\Rightarrow$ (i).

Suppose that condition (i) is satisfied. Let $g: X\rightarrow X$ be the identity mapping. By hypothesis, there exists a continuous mapping $g': Y\rightarrow X$ such that $g'\circ f=g$ . This implies that the mapping $f$ is injective and that $g'$ is a continuous retraction of $f$ .

The equivalence of the properties (ii) and (iii) is immediate.

Definition 5 01VI

Let X be a topological space and let A be a subset of X. A contraction of X onto A is a homotopy $\sigma : X\times \mathbf{I}\rightarrow X$ fixed on A whose origin is the identity mapping of X and whose term is a retraction of X onto A. If there exists a contraction of X onto A, one says that X can be contracted onto A, or again that A is a retractile subset of X.

Remark 01VJ

In other words, a contraction of X onto A is a homotopy $\sigma : X\times \mathbf{I}\rightarrow X$ having the following properties:

(i) $\sigma (x,0) =x$ for every $x\in X$ ;

(ii) $\sigma (x,1)\in A$ for every $x\in X$ ;

(iii) $\sigma (x, t) =x$ for every $x\in A$ and every $t\in \mathbf{I}$ .

For there to exist a contraction of X onto A, it is necessary and sufficient that the canonical injection of A into X be a homotopy of the pair $(A,A)$ onto the pair $(X,A)$ .

Let $a$ be a point of X. A contraction of X onto the subspace $\{a\}$ is nothing other than a homotopy pointed at $a$ joining the mapping Id $_X$ to the constant mapping with image $\{a\}$ . Consequently, X contracts onto $\{a\}$ if and only if X is contractible at $a$ (Example 3 of III, p. 234).

Let X be a topological space and let A be a subset of X. Let $\sigma$ be a contraction of X onto A. The mapping $r$ of X into A defined by $r(x) =\sigma (x,1)$ is a retraction of X onto A and $\sigma$ is a homotopy joining Id $_X$ to $r$ . The relations $r\circ i=$ Id $_A$ and $i\circ r=r$ then imply that the mappings $i$ and $r$ are homotopies, inverse to one another up to homotopy.

Definition 6 01VK

With the preceding notations, one says that $\sigma$ is a strong contraction if, moreover, one has $r(\sigma (x, t)) =r(x)$ for every $x\in X$ and every $t\in \mathbf{I}$ ,

Example 01VL

Let X be the complement of the origin in $\mathbf{B}_n$ . The mapping of $X\times \mathbf{I}$ into X given by $(x, t)\mapsto ((1-t) +t\frac{1}{\|x\|})x$ is a strong contraction of X onto $\mathbf{S}_{n-1}$ . The retraction of X onto $\mathbf{S}_{n-1}$ associated with it is the mapping given by $x\mapsto x/\|x\|$ .

Lemma 2 01VM

Let X be a topological space, let U be an open subset of X and let $\sigma$ be a strong contraction of X onto X-U. Then $\sigma (U\times \mathbf{I})\subset \overline{U}$ . In particular, $\sigma (U\times \{1\})$ is contained in the boundary of U.

Let $x\in U$ . The set of real numbers $t\in \mathbf{I}$ such that $\sigma (x, t)\in U$ is open in $\mathbf{I}$ and contains 0; let $s$ be its least upper bound. We have $\sigma (x, s)\in \overline{U}$ . If $s <1,\sigma (x, s)\notin U$ , by definition of $s$ ; the same is true if $s= 1$ since $\sigma (x,1)\in X-U$ . Consequently, $\sigma (x, s)\in$ Fr(U). By definition of a strong contraction, we then have $\sigma (x, s) =\sigma (\sigma (x, s),1) =\sigma (x,1)$ ; in particular, $\sigma (x,1)\in \overline{U}$ . Consequently, $\sigma (x,1)$ belongs to the boundary $\overline{U}\cap (X-U)$ of U.

Soit $t\in \mathbf{I}$ ; si $\sigma (x, t)\notin U$ , il vient encore $\sigma (x, t) =\sigma (\sigma (x, t),1) =$ $\sigma (x,1)$ , donc $\sigma (x, t)\in \overline{U}$ , d’où le lemme.

6. Cylinder of a mapping

Let X and Y be topological spaces and let $f$ be a continuous mapping of X into Y. Denote by U the topological space sum of the topological space $U_1= X\times \mathbf{I}$ and of the space $U_2$ = Y and identify $U_1$ and $U_2$ with subspaces of U by the canonical injections. Let R be the equivalence relation on U the finest for which the points $(x,1)$ of $U_1$ and $f(x)$ of $U_2$ are equivalent, for every $x\in X$ .

Definition 7 01VN

The cylinder of the mapping $f$ is called and is denoted by Cyl( $f$ ), the topological space quotient $U/R$ .

Denote by $\alpha_f: X\times \mathbf{I}\rightarrow$ Cyl( $f$ ) and $\beta_f: Y\rightarrow$ Cyl( $f$ ) the restrictions to $U_1$ and $U_2$ of the canonical surjection of U onto $U/R$ . The mapping $\alpha_f$ is a homotopy and its term is $\beta_f\circ f$ .

Proposition 4 (Universal Property of Cylinders)

Let Z be a topological space, let $\beta : Y\rightarrow Z$ be a continuous mapping and let $\alpha : X\times \mathbf{I}\rightarrow Z$ be a homotopy whose term is $\beta \circ f$ . There exists a unique continuous mapping $\varphi$ from Cyl( $f$ ) into Z such that $\alpha =\varphi \circ \alpha_f$ and $\beta =\varphi \circ \beta_f$ .

The mapping $\psi$ from U into Z which coincides with $\alpha$ in $X\times \mathbf{I}$ and with $\beta$ in Y is continuous. We have $\alpha (x,1) =\beta (f(x))$ for every $x\in X$ , hence $\psi$ defines by passing to the quotient a continuous mapping $\varphi$ from Cyl( $f$ ) into Z such that $\alpha =\varphi \circ \alpha_f$ and $\beta =\varphi \circ \beta_f$ . Since the images of $\alpha_f$ and $\beta_f$ cover Cyl( $f$ ) $,\varphi$ is the only mapping from Cyl( $f$ ) into Z satisfying these relations.

Example 1 01VO

Let $X',Y'$ be topological spaces and let $f': X'\rightarrow Y'$ be a continuous mapping. Let $u: X\rightarrow X'$ and $v: Y\rightarrow Y'$ be continuous mappings such that $f'\circ u=v\circ f$ . There exists a unique continuous mapping $w:$ Cyl( $f$ ) $\rightarrow$ Cyl( $f'$ ) such that $w\circ \alpha_f=\alpha_{f'}\circ (u\times$ Id $_{\mathbf{I}})$ and $w\circ \beta_f=\beta_{f'}\circ v$ .

According to prop. 4, applied to Z = Y and $\beta =$ Id $_Y$ , there exists a unique continuous mapping $\gamma_f:$ Cyl( $f$ ) $\rightarrow Y$ such that $\gamma_f(\alpha_f(x, s)) =f(x)$ and $\gamma_f(\beta_f(y)) =y$ for $x\in X,s\in \mathbf{I}$ and $y\in Y$ .

Proposition 5 01VP

The mapping $\alpha_f$ induces a homeomorphism of $X\times [0,1[$ onto an open subset of Cyl( $f$ ). The mapping $\beta_f$ defines a homeomorphism of Y onto the complement of this open subset. The mapping $\beta_f\circ \gamma_f$ is a continuous retraction of Cyl( $f$ ) onto $\beta_f(Y)$ .

The set $X\times [0,1[$ is an open subset of U, saturated for the relation R, and the equivalence relation induced by R in $X\times [0,1[$ is the equality relation. The first assertion follows from this (TG, I, p. 23, cor. 10).

The complement of $\alpha_f(X\times [0,1[)$ is $\beta_f(Y)$ . Since one has $\gamma_f\circ \beta_f=$ Id $_Y,\beta_f$ defines a homeomorphism of Y onto $\beta_f(Y)$ and $\beta_f\circ \gamma_f$ is a continuous retraction of the canonical injection of $\beta_f(Y)$ into Cyl( $f$ ).

The closed subspace $\beta_f(Y)$ of Cyl( $f$ ) is called the basis of the cylinder of $f$ . The mapping $\beta_f\circ \gamma_f$ is called the canonical retraction of Cyl( $f$ ) onto its basis.

Consider the mappings $\sigma_1: U_1\times \mathbf{I}\rightarrow$ Cyl( $f$ ) and $\sigma_2: U_2\times \mathbf{I}\rightarrow$ Cyl( $f$ ) defined by

$\sigma_1((x, s), t) =\alpha_f(x,(1-t)s+t)$ for $(x, s)\in X\times \mathbf{I}$ and $t\in \mathbf{I}$

$\sigma_2(y, t) =\beta_f(y)$ for $y\in Y$ and $t\in \mathbf{I}$ .

They are continuous. For $x\in X$ and $t\in \mathbf{I}$ , one has

\sigma_1((x,1), t) =\alpha_f(x,1) =\beta_f(f(x)) =\sigma_2(f(x), t)

There therefore exists a unique mapping $\sigma_f$ of Cyl( $f$ ) $\times \mathbf{I}$ into Cyl( $f$ ) such that $\sigma_f\circ (\alpha_f\times$ Id $_{\mathbf{I}}) =\sigma_1$ and $\sigma_f\circ (\beta_f\times$ Id $_{\mathbf{I}}) =\sigma_2$ . The mapping $\sigma_f$ is continuous (I, p. 19, prop. 10).

Proposition 6 01VQ

The mapping $\sigma_f:$ Cyl( $f$ ) $\times \mathbf{I}\rightarrow$ Cyl( $f$ ) is a strong contraction of Cyl( $f$ ) onto its base. Its term is the canonical retraction of Cyl( $f$ ) onto its base.

The mapping $\sigma_f$ is a homotopy. The relations $\sigma_f(\alpha_f(x, s),0) =$ $\alpha_f(x, s)$ and $\sigma_f(\beta_f(y),0) =\beta_f(y)$ imply that the origin of $\sigma_f$ is the identity mapping of Cyl( $f$ ). Let $r_f$ denote the term of $\sigma_f$ . The relations

\sigma_f(\alpha_f(x, s),1) =\alpha_f(x,1) =\beta_f(f(x)) = (\beta_f\circ \gamma_f)(\alpha_f(x, s))

and $\sigma_f(\beta_f(y),1) =\beta_f(y) = (\beta_f\circ \gamma_f)(\beta_f(y))$ imply that $r_f$ is the canonical retraction $\beta_f\circ \gamma_f$ of Cyl( $f$ ) onto its base.

For $(x, s)\in X\times \mathbf{I}$ and $t\in \mathbf{I}$ , we have

r_f(\sigma_f(\alpha_f(x, s), t)) =r_f(\alpha_f(x, s(1-t) +t)) =\beta_f(f(x)) =r_f(\alpha_f(x, s))

For $y\in Y$ and $t\in \mathbf{I}$ , we have $r_f(\sigma_f(\beta_f(y), t)) =r_f(\beta_f(y))$ . Consequently, $\sigma_f$ is a strong contraction of Cyl( $f$ ) onto its base.

The mapping $\sigma_f$ is called the canonical contraction of the cylinder of $f$ onto its base.

Remark 1 01VR

Let A be a subset of $X\times \mathbf{I}$ and let $A_1$ be the set of points $x\in X$ such that $(x,1)\in A$ . We have $\alpha^{-1}_f(\alpha_f(A)) = A\cup \overset{-1}{f}(f(A_1))\times \{1\}$ and $\beta^{-1}_f(\alpha_f(A)) =f(A_1)$ . It follows that the mapping $\alpha_f$ is closed (resp. open) if $f$ is.

Remark 2 01VS

Suppose that the mapping $f$ is proper. Let P be a point of Cyl( $f$ ). If P = $\alpha_f(x, t)$ , with 0 $\leqslant t <$ 1 and $x\in$ X, we have $\alpha^{-1}_f(P) = (x, t)$ . In the contrary case, there exists $y\in Y$ such that $P =\beta_f(y)$ and $\alpha^{-1}_f(P) =\overset{-1}{f}(y)\times \{1\}$ . This proves that the fibres of the mapping $\alpha_f$ are quasi-compact. The mapping $\alpha_f$ is then proper (TG, I, p. 75, th. 1), since it is closed.

By prop. 5 and TG, I, p. 72, prop. 2, the mapping $\beta_f$ is itself proper. It follows that, if $f$ is proper, the canonical surjection of U onto Cyl( $f$ ) is proper and hence, in particular, universally strict (I, p. 20, corollary).

Remark 3 01VT

If the spaces X and Y are separated, the same is true of the mapping cylinder of the mapping $f$ . Indeed, let $z$ and $z'$ be two distinct points of Cyl( $f$ ) and let us prove that they possess disjoint neighbourhoods. Distinguish three cases.

– There exist $(x, t)$ and $(x', t')\in X\times [0,1[$ such that $z=\alpha_f(x, t)$ and $z'=\alpha_f(x', t')$ .

In this case, the assertion follows from the fact that the space $X\times [0,1[$ is separated (TG, I, p. 54, prop. 7) and that the mapping $\alpha_f$ induces a homeomorphism of this space onto an open subspace of Cyl( $f$ ).

– There exists $(x, t)\in X\times [0,1[$ such that $z=\alpha_f(x, t)$ and $y'\in Y$ such that $z'=\beta_f(y')$ .

Then, $\alpha_f(X\times [0,\frac{t+1}{2}[)$ and Cyl( $f$ ) $-\alpha_f(X\times [0,\frac{t+1}{2}])$ are disjoint open neighbourhoods of $z$ and $z'$ in Cyl( $f$ ).

– There exist $y$ and $y'\in Y$ such that $z=\beta_f(y),z'=\beta_f(y')$ .

In this case, $y=\not y'$ ; since Y is separated, there exists an open neighbourhood V of $y$ in Y and an open neighbourhood $V'$ of $y'$ in Y such that $V\cap V'=\emptyset$ . Then, $(\beta_f\circ \gamma_f)^{-1}(V)$ and $(\beta_f\circ \gamma_f)^{-1}(V')$ are disjoint open subsets of Cyl( $f$ ) containing respectively $y$ and $y'$ .

7. The homotopy extension property

Definition 8 01VU

Let X be a topological space and let A be a subset of X. We say that the pair $(X,A)$ possesses the homotopy extension property if, for every topological space Y, every continuous mapping $f: X\rightarrow Y$ and every homotopy $\sigma : A\times \mathbf{I}\rightarrow Y$ whose term is the mapping $f|A$ , there exists a homotopy $\tau : X\times \mathbf{I}\rightarrow Y$ extending $\sigma$ and whose term is the mapping $f$ .

Remark 1 01VV

Let X be a topological space and let A be a subset of X such that the pair $(X,A)$ possesses the homotopy extension property. Let Y be a topological space, let $f: X\rightarrow Y$ be a continuous mapping and let $\sigma : A\times \mathbf{I}\rightarrow Y$ be a homotopy whose origin is the mapping $f|A$ . The mapping $\overline{\sigma}: A\times \mathbf{I}\rightarrow Y$ defined by $(a, t)\mapsto \sigma (a,1-t)$ is a homotopy with term $f|A$ ; let $\tau : X\times \mathbf{I}\rightarrow Y$ be a homotopy with term $f$ which extends $\overline{\sigma}$ . The mapping $\overline{\tau}: X\times \mathbf{I}\rightarrow Y$ given by $(x, t)\mapsto \tau (x,1-t)$ is then a homotopy which extends $\sigma$ and whose origin is the mapping $f$ .

Let X be a topological space, let A be a subspace of X and let $i: A\rightarrow X$ be the canonical injection. Let $\alpha_i: A\times \mathbf{I}\rightarrow$ Cyl( $i$ ) and $\beta_i: X\rightarrow$ Cyl( $i$ ) denote the canonical mappings. Let $j:$ Cyl( $i$ ) $\rightarrow X\times \mathbf{I}$ be the unique continuous mapping such that $j(\alpha_i(a, s)) = (i(a), s)$ and $j(\beta_i(x)) = (x,1)$ for $a\in A,s\in \mathbf{I}$ and $x\in X$ . It is injective; its image is the subspace $(A\times \mathbf{I})\cup (X\times \{1\})$ of $X\times \mathbf{I}$ . The mapping $j$ is closed if A is closed in X. It is not always strict (III, p. 325, exerc. 17).

Proposition 7 01VW

With the above notations, the following assertions are equivalent:

(i) The pair $(X,A)$ has the homotopy extension property;

(ii) For every topological space Y and every continuous mapping $g:$ Cyl( $i$ ) $\rightarrow Y$ , there exists a continuous mapping $g': X\times \mathbf{I}\rightarrow Y$ such that $g=g'\circ j$ ;

(iii) The injection $j$ has a continuous retraction;

(iv) The mapping $j$ is strict and there exists a contraction of $X\times \mathbf{I}$ onto the image of $j$ .

Suppose that the pair $(X,A)$ has the homotopy extension property. Let $g:$ Cyl( $i$ ) $\rightarrow Y$ be a continuous mapping; put $\sigma =g\circ \alpha_i$ and $f=g\circ \beta_i$ . The mapping $f: X\rightarrow$ Y is continuous and $\sigma : A\times \mathbf{I}\rightarrow Y$ is a homotopy whose end is the mapping $f|A$ . There therefore exists a homotopy $g': X\times \mathbf{I}\rightarrow Y$ with end $f$ extending $\sigma$ . We have $g'\circ j(\alpha_i(a, s)) =g'(a, s) =\sigma (a, s) =g(\alpha_i(a, s))$ and $g'\circ j(\beta_i(x)) =$ $g'(x,1) =f(x) =g(\beta_i(x))$ for $a\in A,s\in \mathbf{I}$ and $x\in X$ . Consequently, $g'\circ j=g$ , whence (ii).

Conversely, suppose that assertion (ii) is satisfied, and let us prove that the pair $(X,A)$ has the homotopy extension property. Let Y be a topological space, let $f: X\rightarrow Y$ be a continuous mapping and let $\sigma : A\times \mathbf{I}\rightarrow Y$ be a homotopy whose end is equal to $f|A$ . There exists a unique continuous mapping $g:$ Cyl( $i$ ) $\rightarrow Y$ such that $g\circ \alpha_i=\sigma$ and $g\circ \beta_i=f$ (III, p. 238, prop. 4). Any mapping $g': X\times \mathbf{I}\rightarrow Y$ such that $g=g'\circ j$ is then a homotopy with end $f$ extending $\sigma$ .

The equivalence of assertions (ii) and (iii) is a particular case of lemma 1 of III, p. 235.

Suppose that the injection $j:$ Cyl( $i$ ) $\rightarrow X\times \mathbf{I}$ admits a continuous retraction $r$ . The mapping $j$ therefore defines a homeomorphism of Cyl( $i$ ) onto the subspace $T = (A\times \mathbf{I})\cup (X\times \{1\})$ of $X\times \mathbf{I}($ loc. cit.). It is therefore strict.

Put $\rho =$ pr $_1\circ j\circ r$ and $\theta =$ pr $_2\circ j\circ r$ . If $x\in A$ or if $s= 1$ , we have $\rho (x, s) =x$ and $\theta (x, s) =s$ For $(x, s)\in X\times \mathbf{I}$ and $t\in \mathbf{I}$ , put

\sigma ((x, s), t) = (\rho (x,(1-t) +st),(1-t)s+t\theta (x, s))

The mapping $\sigma : (X\times \mathbf{I})\times \mathbf{I}\rightarrow X\times \mathbf{I}$ is continuous. For $(x, s)\in X\times \mathbf{I}$ , we have

\sigma ((x, s),0) = (\rho (x,1), s) = (x, s)

and

\sigma ((x, s),1) = (\rho (x, s), \theta (x, s)) =j\circ r(x, s)

for $x\in X$ and $t\in \mathbf{I}$ , we have

\sigma ((x,1), t) = (\rho (x,1),(1-t) +t\theta (x,1)) = (x,1)

whereas, for $(x, s)\in A\times \mathbf{I}$ and $t\in \mathbf{I}$ , we have

\sigma ((x, s), t) = (x,(1-t)s+ts) = (x, s)

Hence $\sigma$ is a contraction of $X\times \mathbf{I}$ onto T. This shows that assertion (iii) implies assertion (iv).

The implication (iv) $\Rightarrow$ (iii) is obvious.

Remark 2 01VX

Let X be a separated topological space and let A be a subset of X such that the pair $(X,A)$ has the homotopy extension property. Let us denote by $i: A\rightarrow$ X the canonical injection. Let $j:$ Cyl( $i$ ) $\rightarrow X\times \mathbf{I}$ be the canonical injection and let $r$ be a continuous retraction of the mapping $j$ . The subspace $j$ (Cyl( $i$ )) is equal to the set of pairs $(x, t)\in X\times \mathbf{I}$ such that $j(r(x, t)) = (x, t)$ . Since the space $X\times \mathbf{I}$ is separated, the subset $j$ (Cyl( $i$ )) is closed in $X\times \mathbf{I}$ . The set A, equal to the set of $x\in X$ such that $(x,0)\in j$ (Cyl( $i$ )), is then a closed subset of X.

Lemma 3 01VY

Let X and Y be topological spaces, let $p: X\rightarrow Y$ be a continuous proper and open mapping, and let $f: X\rightarrow \mathbf{R}$ be a continuous mapping. The mapping $g: Y\rightarrow \overline{\mathbf{R}}$ given by $y\mapsto$ sup $_{x\in\overset{-1}{p}(y)}f(x)$ is continuous.

Let $b\in Y$ . Since $p$ is proper, its fibre $\overset{-1}{p}(b)$ is a quasi-compact space; hence we have $g(b)\in \mathbf{R}\cup \{-\infty \}$ .

Let $m\in \mathbf{R}$ such that $g(b)< m$ . For every $a\in \overset{-1}{p}(b)$ , one has $f(a)\leqslant$ $g(b)< m$ ; let $V_a$ be a neighbourhood of $a$ in X such that $f(x)< m$ for every $x\in V_a$ . The union V of the sets $V_a$ is a neighbourhood of $\overset{-1}{p}(b)$ in X. By Lemma 5 (I, p. 75), there exists a neighbourhood W of $b$ in Y such that $\overset{-1}{p}(W)\subset V$ . For every $y\in W$ , one has $g(y)\leqslant m$ . This proves that $g$ is upper semi-continuous at $b$ .

Let us now prove that $g$ is lower semi-continuous at $b$ . We may suppose that $g(b)\in \mathbf{R}$ . Let $m\in \mathbf{R}$ such that $m < g(b)$ . Let $a\in \overset{-1}{p}(b)$ such that $m < f(b)$ ; let then V be a neighbourhood of $a$ in X such that $f(x)> m$ for every $x\in V$ . It follows that $g(y)> m$ for every $y\in p(V)$ . Since $p$ is open, $p(V)$ is a neighbourhood of $b$ , so that $g$ is lower semi-continuous.

The lemma is thus proved.

Theorem 1 01VZ

Let X be a topological space and let A be a closed subspace of X; denote by $i: A\rightarrow X$ the canonical injection. The following assertions are equivalent:

(i) The pair $(X,A)$ has the homotopy extension property ;

(ii) There exists a continuous mapping $\varphi : X\rightarrow \mathbf{I}$ such that $A =\overset{-1}{\varphi}(0)$ and a homotopy $\sigma : X\times \mathbf{I}\rightarrow X$ fixed on A whose term is the identity mapping of X and such that $\sigma (x,0)\in A$ for every point $x\in X$ such that $\varphi (x)= 1\not$ .

(iii) There exists a continuous mapping $\varphi : X\rightarrow \mathbf{R}_+$ such that $A =\overset{-1}{\varphi}(0)$ and a homotopy $\sigma :\overset{-1}{\varphi}(\mathbf{I})\times \mathbf{I}\rightarrow X$ , fixed on A, such that $\sigma (x,1) =x$ and $\sigma (x,0)\in A$ for every $x\in \overset{-1}{\varphi}(\mathbf{I})$ .

(iv) There exists a continuous mapping $\varphi : X\rightarrow \mathbf{R}_+$ such that $A\subset \overset{-1}{\varphi}(0)$ and a continuous mapping

\sigma :\{(x, t)\in X\times \mathbf{I}|t+\varphi (x)\geqslant 1\} \rightarrow X

such that $\sigma (x,1) =x$ for every $x\in X$ and $\sigma (x,1-\varphi (x))\in A$ for every $x\in X$ such that $\varphi (x)\leqslant 1$ .

We shall denote by $i$ the canonical injection of A into X $,\alpha_i: A\times \mathbf{I}\rightarrow$ Cyl( $i$ ) and $\beta_i: X\rightarrow$ Cyl( $i$ ) the canonical mappings, and by $j$ the mapping of Cyl( $i$ ) into $X\times \mathbf{I}$ such that $j(\alpha_i(x, s)) = (x, s)$ for $(x, s)\in A\times \mathbf{I}$ and $j(\beta_i(x)) = (x,1)$ for $x\in X$ .

Suppose that the pair $(X,A)$ has the homotopy extension property and let $r: X\times \mathbf{I}\rightarrow$ Cyl( $i$ ) be a continuous retraction of the mapping $j($ III, p. 241, Proposition 7). Let us denote by $\sigma$ the continuous mapping pr $_1\circ j\circ r$ of $X\times \mathbf{I}$ into X. For every $(x, t)\in X\times \mathbf{I}$ , one has $|$ pr $_2(j(r(x, t)))-t|\leqslant 1$ . Since $\mathbf{I}$ is compact, the first projection pr $_1: X\times \mathbf{I}\rightarrow X$ is a proper mapping; it is also open. By Lemma 3, the mapping $\varphi : X\rightarrow \mathbf{I}$ given by

$x\mapsto$ sup $_{t\in\mathbf{I}}(|$ pr $_2(j(r(x, t)))-t|)$

is therefore continuous.

For $x\in X$ , one has $\sigma (x,1) =x$ . Let $x\in X$ be such that $\sigma (x,0)\notin A$ ; then pr $_2(j(r(x,0))) = 1$ and $\varphi (x)\geqslant 1$ . For $x\in A$ and $t\in \mathbf{I}$ , one has $j(r(x, t)) = (x, t)$ ; consequently, $\sigma (x, t) =x$ and $\varphi (x) = 0$ . Conversely, let $x\in X$ be such that $\varphi (x) = 0$ . One has therefore pr $_2(j(r(x, t))\leqslant t$ for every $t\in \mathbf{I}$ ; if $t <1$ , this implies $j(r(x, t))\in A\times \mathbf{I}$ ; since A is closed in X, one has therefore $j(r(x,1))\in A\times \mathbf{I}$ , hence $x\in A$ .

This proves that (i) implies (ii).

Let $\varphi$ and $\sigma$ be mappings satisfying the properties of assertion (ii). Put $\varphi_1= 2\varphi$ and let $\sigma_1$ be the restriction of $\sigma$ to $\overset{-1}{\varphi}_{1}(\mathbf{I})\times \mathbf{I}$ . We have $\overset{-1}{\varphi}_{1}(0) = A$ ; for $a\in A,\sigma_1(a, t) =a$ for all $a\in A$ . Let $x\in X$ be such that $\varphi_1(x)\leqslant 1$ ; we have $\sigma_1(x,1) =x$ ; moreover, $\varphi (x) =\varphi_1(x)/2<1$ , hence $\sigma_1(x,0)\in A$ . Thus, (ii) implies (iii).

Let us prove that (iii) implies (iv). Let $\varphi : X\rightarrow \mathbf{R}$ and $\sigma :\overset{-1}{\varphi}(\mathbf{I})\times$ $\mathbf{I}\rightarrow X$ be as in the statement; put $B =\overset{-1}{\varphi}(\mathbf{I})$ and $C =\overset{-1}{\varphi}([1,+\infty [)$ .

Let $u_1$ be the mapping from $B\times \mathbf{I}$ into X such that $u_1(x, t) =\sigma (x,1-$ $(1-t)/2\varphi (x))$ if $t+ 2\varphi (x)\geqslant 1$ and $\varphi (x)>0$ and $u_1(x, t) =\sigma (x,0)$ otherwise. By the following lemma 4, it is continuous.

Let $u_2$ be the mapping from $B\times \mathbf{I}$ into X such that $u_2(x, t) =$ $\sigma (x$ , sup(0 $,1-2(1-t)(1-\varphi (x))$ )). It is continuous. For $x\in B$ such that $\varphi (x) = 1/2$ and $t\in \mathbf{I}$ , we have $u_1(x, t) =\sigma (x, t) =u_2(x, t)$ . Let $u: B\times \mathbf{I}\rightarrow X$ denote the mapping such that $u(x, t) =u_1(x, t)$ if $\varphi (x)\leqslant 1/2$ and $u(x, t) =u_2(x, t)$ if $1/2< x\leqslant 1$ ; it is continuous since its restrictions to the closed subspaces $\overset{-1}{\varphi}([0,1/2])\times \mathbf{I}$ and $\overset{-1}{\varphi}([1/2,1])\times \mathbf{I}$ of $B\times \mathbf{I}$ are continuous (TG, I, p. 19, prop. 4).

For $x\in X$ such that $\varphi (x) = 1$ and $t\in \mathbf{I}$ , we have $u(x, t) =u_2(x, t) =$ $\sigma (x,1) =x$ . There therefore exists a unique mapping $\tau : X\times \mathbf{I}\rightarrow X$ which coincides with $u$ in $B\times \mathbf{I}$ and with the mapping pr $_1$ in $C\times \mathbf{I}$ ; it is continuous (loc. cit.).

Let $x\in X$ ; it is verified that one has $\tau (x,1) = 1$ . If, moreover, $x\in A$ , then one has $\varphi (x) = 0$ , hence $\tau (x, t) =u_1(x, t) =\sigma (x,0) =x$ . Finally, if $2\varphi (x)\leqslant 1$ , then $\varphi (x)\leqslant 1/2$ , hence $\tau (x,1-2\varphi (x)) =\sigma (x,0)\in A$ .

This proves that assertion (iv) is satisfied.

Suppose finally that assertion (iv) is satisfied, and let us prove that the pair $(X,A)$ has the homotopy extension property.

Let $C_1$ (resp. $C_2$ ) be the set of pairs $(x, t)\in X\times \mathbf{I}$ such that $t+\varphi (x)\leqslant 1$ (resp. $t+\varphi (x)\geqslant$ 1). These are closed sets. For $(x, t)\in C_1$ , one has $\sigma (x,1-\varphi (x))\in A$ ; let then $\rho_1: C_1\rightarrow$ Cyl( $i$ ) be the mapping defined by $(x, t)\mapsto \alpha_i(\sigma (x,1-\varphi (x)), t+\varphi (x))$ ; it is continuous. Let also $\rho_2: C_2\rightarrow$ Cyl( $i$ ) be the continuous mapping defined by $(x, t)\mapsto \beta_i(\sigma (x, t))$ . For $(x, t)\in C_1\cap C_2$ , one has $t+\varphi (x) = 1$ , hence

\rho_1(x, t) =\alpha_i(\sigma (x,1-\varphi (x)),1) =\beta_i(\sigma (x,1-\varphi (x)) =\rho_2(x, t)

There therefore exists a unique mapping $\rho : X\times \mathbf{I}\rightarrow$ Cyl( $i$ ) which coincides with $\rho_1$ in $C_1$ and with $\rho_2$ in $C_2$ ; it is continuous (TG, I, p. 19, prop. 4).

For $x\in A$ and $t\in \mathbf{I}$ , one has $\varphi (x) = 0$ , hence $t+\varphi (x)\leqslant 1$

\rho (j(\alpha_i(x, t))) =\rho (x, t) =\alpha_i(\sigma (x,1), t) =\alpha_i(x, t)

For $x\in X$ , one also has

\rho (j(\beta_i(x))) =\rho (x,1) =\beta_i(\sigma (x,1)) =\beta_i(x)

Since the images of the mappings $\alpha_i$ and $\beta_i$ cover Cyl( $i$ ), it follows that the mapping $\rho$ is a continuous retraction of the mapping $j$ . Consequently, the pair $(X,A)$ has the homotopy extension property (III, p. 241, prop. 7), which concludes the proof of the theorem.

Lemma 4 01W0

Let X and Y be topological spaces, let $\varphi : X\rightarrow \mathbf{R}_+$ be a continuous mapping, and put $A =\overset{-1}{\varphi}(0)$ . Let $\sigma : X\times \mathbf{I}\rightarrow Y$ be a homotopy which is fixed on A. The mapping $\sigma ': X\times \mathbf{I}\rightarrow Y$ which maps $(x, s)$ to $\sigma (x, s/\varphi (x))$ if $s < \varphi (x)$ and to $\sigma (x,1)$ if $s\geqslant \varphi (x)$ is continuous.

The mapping $\sigma '$ is continuous at every point of the closed subset $\overset{-1}{\varphi}([1,+\infty [)\times \mathbf{I}$ ; it is therefore enough to prove that its restriction to $\overset{-1}{\varphi}(\mathbf{I})\times \mathbf{I}$ is continuous. We may therefore suppose that $\varphi (X)\subset \mathbf{I}$ . Let C and $C'$ be the subspaces of $X\times \mathbf{I}$ consisting of pairs $(x, s)$ such that $s\leqslant \varphi (x)$ and $s\geqslant \varphi (x)$ respectively. They are closed and cover $X\times \mathbf{I}$ . The mapping $\sigma '$ is continuous on $C'$ ; let us prove that it is continuous on C.

Let $\alpha : X\times \mathbf{I}\rightarrow X\times \mathbf{I}$ be the continuous mapping defined by $\alpha (x, t) =$ $(x, t\varphi (x))$ . Its image is equal to C and $\sigma '\circ \alpha$ is the continuous mapping $\sigma$ . Let us prove that $\alpha$ is a proper map. Consider in fact an ultrafilter $\mathfrak{U}$ on $X\times \mathbf{I}$ and a point $(x, t)\in X\times \mathbf{I}$ adherent to the ultrafilter basis $\alpha (\mathfrak{U})$ . Since pr $_1\circ \alpha =$ pr $_1$ , the ultrafilter basis pr $_1(\mathfrak{U})$ on X converges to $x$ . Since $\mathbf{I}$ is compact, there exists a point $s\in \mathbf{I}$ such that the ultrafilter basis pr $_2(\mathfrak{U})$ converges to $s$ . Then $\mathfrak{U}$ converges to $(x, s)$ . Since $\alpha$ is continuous, the ultrafilter basis $\alpha (\mathfrak{U})$ converges to $(x, s\varphi (x))$ . Since $\mathbf{I}$ is separated, we have $s\varphi (x) =t$ , whence $\alpha (x, s) = (x, t)$ , so that $\alpha$ is proper (TG, I, p. 75, Theorem 1). It then follows from I, p. 18, example 2 and Proposition 9 that $\sigma '|C$ is continuous.

Corollary 1 01W1

Let X be a normal topological space and let A be a closed subspace of X. Suppose that there exist a neighbourhood V of A in X and a contraction of V onto A, together with a continuous mapping $f: X\rightarrow \mathbf{R}$ such that $\overset{-1}{f}(0) = A$ . Then the pair $(X,A)$ has the homotopy extension property.

Let $\rho : V\times \mathbf{I}\rightarrow V$ be a contraction of V onto A. By definition of a normal space (TG, IX, p. 41, Definition 1), there exists a continuous mapping $g: X\rightarrow \mathbf{I}$ which is 0 on A and 1 at every point of X - V. Let $\varphi : X\rightarrow \mathbf{R}$ be the mapping defined by $\varphi (x) =|f(x)|+g(x)$ for $x\in X$ ; it is continuous. We have $\overset{-1}{\varphi}(0) = A$ and $\overset{-1}{\varphi}(\mathbf{I})\subset V$ . Let $\sigma$ be the mapping from $\overset{-1}{\varphi}(\mathbf{I})\times \mathbf{I}$ into X defined by $\sigma (x, t) =\rho (x,1-t)$ for $x\in \overset{-1}{\varphi}(\mathbf{I})$ and $t\in \mathbf{I}$ . For $x\in \overset{-1}{\varphi}(\mathbf{I})$ , we have $\sigma (x,1) =\rho (x,0) =x$ and $\sigma (x,0) =\rho (x,1)\in A$ .

The mappings $\varphi$ and $\sigma$ satisfy the conditions of assertion (iii) of theorem 1 of III, p. 243; hence the pair $(X,A)$ possesses the homotopy extension property.

Example 1 01W2

Take for space X the ball $\mathbf{B}_n$ and for subspace A the sphere $\mathbf{S}_{n-1}$ . If V = X $-\{0\}$ , there exists a strong contraction of V onto $\mathbf{S}_{n-1}($ III, p. 237, example). Hence the pair $(\mathbf{B}_n,\mathbf{S}_{n-1})$ possesses the homotopy extension property.

Corollary 2 01W3

Let X and Y be topological spaces, let A be a closed subspace of X, let B be a closed subspace of Y. If the pairs $(X,A)$ and $(Y,B)$ possess the homotopy extension property, the same is true of the pair $(X\times Y,(X\times B)\cup (A\times Y))$ .

Let $\varphi : X\rightarrow \mathbf{R}_+$ and $\sigma :\overset{-1}{\varphi}(\mathbf{I})\times \mathbf{I}\rightarrow$ X, resp. $\varphi ': Y\rightarrow \mathbf{R}_+$

and $\sigma ':\overset{-1}{\psi}(\mathbf{I})\times \mathbf{I}\rightarrow Y$ , satisfy the conditions of assertion (iv) of theorem 1 for the pair $(X,A)$ , resp. for the pair $(Y,B)$ . Let $\psi : X\times Y\rightarrow \mathbf{R}_+$ be the mapping given by $(x, y)\mapsto$ inf( $\varphi (x), \varphi '(x)$ ) ; it is continuous; one has also $\psi (x, y) = 0$ for every $(x, y)\in X\times Y$ such that $x\in A$ or $y\in B$ . For $(x, y, t)\in X\times Y\times \mathbf{I}$ such that $t+\psi (x)\geqslant 1$ , one has $t+\varphi (x)\geqslant 1$ and $t+\varphi '(x)\geqslant 1$ . Thus one defines a continuous mapping

\tau :\{(x, y, t)\in X\times Y\times \mathbf{I}|t+\psi (x)\geqslant 1\} \rightarrow X\times Y

by putting $\tau (x, y, t) = (\sigma (x, t), \sigma '(y, t))$ . For every $(x, y)\in X\times Y$ , one has $\tau (x, y,1) = (\sigma (x,1), \sigma '(y,1)) = (x, y)$ . If, moreover, $\psi (x, y)\leqslant 1$ , then $\varphi (x)\leqslant 1$ or $\varphi '(y)\leqslant 1$ , so that $\sigma (x,1)\in A$ or $\sigma '(y,1)\in B$ ; it follows that $\tau (x, y,1-\psi (x))$ belong to $(A\times Y)\cup (X\times B)$ . This verifies assertion (iv) of theorem 1, whence the corollary.

Example 2 01W4

Here are other cases in which the pair $(X,A)$ possesses the homotopy extension property.

(i) The space X is a paracompact differentiable manifold of class $C^1$ and A a closed submanifold of X. In view of corollary1, this follows from the fact that X is perfectly normal (IX, p. 103, exerc. 11) and that A possesses a tubular neighbourhood in X;

(ii) The space X is a cellular space and A a full subspace with respect to a given cellular decomposition of X.

8. Attachment of a topological space

Let $X,B$ be topological spaces, let A be a subspace of B, and let $f: A\rightarrow X$ be a continuous mapping. Let Y denote the topological sum of the spaces $Y_1= X$ and $Y_2= B$ , and identify $Y_1$ and $Y_2$ with subspaces of Y by the canonical injections. Let R be the finest equivalence relation in Y for which the elements $a$ of $Y_2$ and $f(a)$ of $Y_1$ are equivalent, for every $a\in A$ . The relation R is the equality relation in $Y_1$ . Let $x\in Y_1$ and let $b\in Y_2$ ; one has $xRb$ if and only if $b\in A$ and $f(b) =x$ . Let $b, b'\in Y_2$ ; in order that one have $bRb'$ , it is necessary and sufficient that $b=b'$ , or that $b, b'\in A$ and $f(b) =f(b')$ .

Definition 9 01W5

The quotient topological space $Y/R$ is called the space obtained by attaching the space B to the space X along the mapping $f$ . It is denoted by $X\cup_fB$ .

Let $\alpha_f: X\rightarrow X\cup_fB$ and $\beta_f: B\rightarrow X\cup_fB$ denote the restrictions to $Y_1$ and $Y_2$ of the canonical surjection of Y onto $Y/R$ . One has $\alpha_f\circ f=\beta_f|A$ .

Remark 1 01W6

The mapping $\alpha_f$ is injective. For every subset U

of X, one has $\alpha^{-1}_f(\alpha_f(U)) = U$ and $\beta^{-1}_f(\alpha_f(U)) =\overset{-1}{f}(U)$ . If A is an open (resp. closed) subset of B, these relations imply that $\alpha_f$ is an open (resp. closed) mapping of X into $X\cup_fB$ .

Let U be an open subset of X and let V be an open subset of B such that $\overset{-1}{f}(U) = V\cap A$ . The union of the subsets U of $Y_1$ and V of $Y_2$ is a saturated open subset of Y; its image in $X\cup_fB$ is therefore open, and its trace on $\alpha_f(X)$ is $\alpha_f(U)$ . It follows that the mapping $\alpha_f$ defines a homeomorphism of X onto its image in $X\cup_fB$ .

Remark 2 01W7

Let V be a subset of B. One has $\alpha^{-1}_f(\beta_f(V)) =f(V\cap A)$ and

$\beta^{-1}_f(\beta_f(V)) = V\cup \overset{-1}{f}(f(V\cap A))$ . If A is closed in B and if the mapping $f$ is closed, the mapping $\beta_f$ is therefore closed. Analogously, if A is open in B and if the mapping $f$ is open, the mapping $\beta_f$ is open.

In these two cases, the mapping $\beta_f$ then induces a homeomorphism of B-A onto its image.

Remark 3 01W8

Suppose that A is closed and that the mapping $f$ is proper. We have just seen that the mapping $\beta_f$ is closed. For every $x\in X$ ,

$\beta^{-1}_f(\alpha_f(x)) =\overset{-1}{f}(x)$ . It is therefore a quasi-compact subset of A (TG,

I, p. 75, Theorem 1), hence also of B. For every $b\in B,\beta^{-1}_f(\beta_f(b))$

equal to $\{b\}$ if $b\in B-A$ , and to $\overset{-1}{f}(f(b))$ if $b\in A$ ; in both cases, it is a quasi-compact subset of B. The fibres of the mapping $\beta_f$ are therefore quasi-compact; consequently (loc. cit.), the mapping $\beta_f$ is proper.

The mapping $\alpha_f$ is likewise proper (TG, I, p. 72, prop. 2). It follows that the canonical mapping of Y onto $X\cup_fB$ is proper.

Example 1 01W9

Let X and Y be topological spaces and let $f: X\rightarrow Y$ be a continuous mapping. The cylinder Cyl( $f$ ) is none other than the space obtained by attaching the space Y to the space $X\times \mathbf{I}$ along the mapping $f\circ$ pr $_1$ of $X\times \{1\}$ into Y.

Example 2 01WA

Let X be a topological space, let B be a topological space and let A be a subspace of B, and let $f: A\rightarrow X$ be a continuous mapping inducing a homeomorphism of A onto its image.

Put $A'=f(A)$ ; let $f'$ be the mapping of $A'$ into B which associates to each $x\in A'$ the unique inverse image of $x$ by $f$ . The mapping $f'\circ f$ is continuous; since $f$ is strict, the mapping $f'$ is continuous. There exists a unique continuous mapping $u$ of the space $X\cup_fB$ into the space $B\cup_{f'}X$ which maps $\alpha_f(x)$ to $\beta_{f'}(x)$ for $x\in X$ and $\beta_f(b)$ to $\alpha_{f'}(b)$ for $b\in B$ ; it is a homeomorphism mapping the subspace $\alpha_f(X)$ onto the subspace $\beta_{f'}(X)$ .

Proposition 8 01WB

Let Z be a topological space.

a) Let $u: X\rightarrow Z$ and $v: B\rightarrow Z$ be continuous mappings such that $v(a) =u(f(a))$ for every $a\in A$ . Then there exists a unique continuous mapping $w: X\cup_fB\rightarrow Z$ such that $w\circ \alpha_f=u$ and $w\circ \beta_f=v$ .

b) Let $\sigma : X\times \mathbf{I}\rightarrow Z$ and $\tau : B\times \mathbf{I}\rightarrow Z$ be homotopies. Suppose that $\tau$ is fixed on A and that $\sigma (f(a), t) =\tau (a, t)$ for every $a\in A$ and every $t\in \mathbf{I}$ . Then there exists a unique homotopy $\eta : (X\cup_fB)\times \mathbf{I}\rightarrow Z$ such that $\eta (\alpha_f(x), t) =\sigma (x, t)$ and $\eta (\beta_f(b), t) =\tau (b, t)$ for $x\in X,b\in B$ and $t\in \mathbf{I}$ . This homotopy is fixed on $\beta_f(A)$ .

The proposition results immediately from the definition of a quotient space and prop. 10 (I, p. 19).

Corollary 1 01WC

Let $B'$ be a topological space, let $A'$ be a subspace of $B'$ , and let $f': A'\rightarrow$ X be a continuous mapping. Let $v: B\rightarrow B'$ be a continuous mapping such that $v(A)\subset A'$ and $f$ = $f'\circ (v|A)$ . Let $w: X\cup_fB\rightarrow X\cup_{f'}B'$ be the unique continuous mapping such that $w\circ \alpha_f=\alpha_{f'}$ and $w\circ \beta_f=\beta_{f'}\circ v$ . If $v$ defines a homotopy equivalence of the pair $(B,A)$ onto the pair $(B',A')$ , then $w$ defines a homotopy equivalence of the pair $(X\cup_fB, \alpha_f(X))$ onto the pair $(X\cup_{f'}B', \alpha_{f'}(X))$ .

Let $v': B'\rightarrow B$ be a continuous mapping, let $\tau : B\times \mathbf{I}\rightarrow B$ be a homotopy fixed on A joining Id $_B$ to $v'\circ v$ , and let $\tau ': B'\times \mathbf{I}\rightarrow B'$ be a homotopy fixed on $A'$ joining Id $_{B'}$ to $v\circ v'$ . For $a'\in A'$ and $a=v'(a')$ , we have the relations $a'=v(a)$ and $\beta_f(v'(a')) =\beta_f(a) =\alpha_f(f(a)) =\alpha_f(f'(a'))$ . By Prop. 8, a), there exists a unique mapping $w': X\cup_{f'}B'\rightarrow$ $X\cup_fB$ such that $w'\circ \alpha_{f'}=\alpha_{f'}$ and $w'\circ \beta_{f'}=\beta_f\circ v'$ . By Prop. 8, b), there exists a unique homotopy $\eta : (X\cup_fB)\times \mathbf{I}\rightarrow X\cup_{f'}B'$ such that $\eta (\alpha_f(x), t) =\alpha_{f'}(x)$ and $\eta (\beta_f(b), t) =\beta_{f'}(\tau (b, t))$ for $x\in X,b\in B$ and $t\in \mathbf{I}$ . Analogously there exists a unique homotopy $\eta ': (X\cup_{f'}B')\times \mathbf{I}\rightarrow$ $X\cup_fB$ such that $\eta '(\alpha_{f'}(x), t) =\alpha_f(x)$ and $\eta '(\beta_{f'}(b), t) =\beta_f(\tau '(b, t))$ for $x\in X,b\in B'$ and $t\in \mathbf{I}$ . The mapping from $(X\cup_fB)\times \mathbf{I}$ into $X\cup_fB$ given by $(x, t)\mapsto \eta '(\eta (x, t), t)$ is then a homotopy fixed on $\alpha_f(X)$ joining the identity mapping of $X\cup_fB$ to the mapping $w'\circ w$ . Analogously, the mapping from $(X\cup_{f'}B')\times \mathbf{I}$ into $X\cup_{f'}B'$ given by $(x, t)\mapsto \eta (\eta '(x, t), t)$ is a homotopy fixed on $\alpha_{f'}(X)$ joining the identity mapping of $X\cup_{f'}B'$ to the mapping $w\circ w'$ . The corollary follows.

Example 3 01WD

Let $i$ be the canonical injection of A into B and let us take for the space $B'$ the cylinder of the mapping $i$ ; let $\alpha_i: A\times \mathbf{I}\rightarrow$ Cyl( $i$ ) and $\beta_i: B\rightarrow$ Cyl( $i$ ) be the canonical mappings and $r_i:$ Cyl( $i$ ) $\rightarrow B$ the canonical retraction of the cylinder Cyl( $i$ ) onto its base. Let $A_0$ be the subspace $\alpha_i(A\times \{0\})$ of Cyl( $i$ ) and let $f_0: A_0\rightarrow X$ be the mapping $f\circ r_i$ . For $a\in A$ , we have

\beta_f\circ r_i(\alpha_i(a,0)) =\beta_f(a) =\alpha_f(f_0(\alpha_i(a,0))) =\alpha_f(f(a))

Let $\eta : X\cup_{f_0}$ Cyl( $i$ ) $\rightarrow X\cup_fB$ be the unique continuous mapping such that $\eta \circ \alpha_{f_0}=\alpha_f$ and $\eta \circ \beta_{f_0}=\beta_f\circ r_i$ . Suppose that the pair $(B,A)$ has the homotopy extension property. Then the mapping $r_i$ defines a homotopy equivalence of the pair (Cyl( $i$ ) $,A_0$ ) with the pair $(B,A)$ (III, p. 243, th. 1). By Corollary 1, the mapping $\eta$ then defines a homotopy equivalence of the pair $(X\cup_{f'}B', \alpha_{f'}(X))$ with the pair $(X\cup_fB, \alpha_f(X))$ . In particular, $\eta$ is a homotopy equivalence.

Corollary 2 01WE

Let $X'$ be a topological space, let $u: X\rightarrow X'$ be a continuous mapping, and put $f'=u\circ f$ . Let $w: X\cup_fB\rightarrow X'\cup_{f'}B$ be the unique continuous mapping such that $w\circ \alpha_f=\alpha_{f'}\circ u$ and $w\circ \beta_f=\beta_{f'}$ . If $u$ defines a homotopy equivalence of the pair $(X, f(A))$ with the pair $(X', f'(A))$ , then $w$ defines a homotopy equivalence of the pair $(X\cup_fB, \beta_f(B))$ with the pair $(X'\cup_{f'}B, \beta_{f'}(B))$ .

The proof is analogous to that of Corollary 1.

Proposition 9 01WF

Let X and B be topological spaces, let A be a subspace of B, and let $f: A\rightarrow X$ be a continuous mapping.

a) If the pair $(B,A)$ has the homotopy extension property, the same is true of the pair $(X\cup_fB, \alpha_f(X))$ .

b) Suppose that the mapping $f$ is injective and strict. If the pair $(X, f(A))$ has the homotopy extension property, the same is true of the pair $(X\cup_fB, \beta_f(B))$ .

a) Suppose that the pair $(B,A)$ has the homotopy extension property. Let Z be a topological space, let $u: X\cup_fB\rightarrow Z$ be a continuous mapping, and let $\sigma :\alpha_f(X)\times \mathbf{I}\rightarrow Z$ be a homotopy whose final term is the restriction of $u$ to the subspace $\alpha_f(X)$ .

Let us put $v_1=u\circ \alpha_f$ and denote by $\tau_1: X\times \mathbf{I}\rightarrow Z$ the mapping given by $(x, t)\mapsto \sigma (\alpha_f(x), t)$ . Put $v_2=u\circ \beta_f$ . Since $\beta_f|A =\alpha_f\circ f$ , the mapping from $A\times \mathbf{I}$ into Z which maps $(a, t)$ to $\sigma (\alpha_f(f(a)), t)$ for $a\in A$ and $t\in \mathbf{I}$ is a homotopy whose final term is equal to the mapping $v_2|A$ . Since the pair $(B,A)$ has the homotopy extension property, there exists a homotopy $\tau_2: B\times \mathbf{I}\rightarrow Z$ whose final term is the mapping $v_2$ and such that $\tau_2(a, t) =\sigma (\alpha_f(f(a)), t)$ for $(a, t)\in A\times \mathbf{I}$ .

For $a\in A$ and $t\in \mathbf{I}$ , we have $\tau_2(a, t) =\tau_1(f(a), t)$ . By Proposition 8, there exists a unique continuous mapping $\sigma : (X\cup_fB)\times \mathbf{I}\rightarrow Z$ such that $\sigma (\alpha_f(x), t) =\sigma_1(x, t)$ and $\sigma (\beta_f(b), t) =\sigma_2(b, t)$ , for $x\in X$ , $b\in B$ and $t\in \mathbf{I}$ . It is a homotopy with final term $u$ which extends $\tau$ , hence assertion a).

b) In view of Example 2 (III, p. 249), assertion b) follows from assertion a).

9. Space obtained by contraction of a subspace

Let X be a topological space and let A be a subset of X. Consider the finest equivalence relation R in X for which all the elements of A are equivalent: two elements of X are equivalent under this relation if they are equal or if they both belong to A.

Definition 10 01WG

The quotient space of X by R is denoted by $X/A$ and is called the space obtained from X by contraction of the subset A.

Let us denote by $\rho : X\rightarrow X/A$ the canonical surjection. Let Y be a topological space and let $f: X\rightarrow Y$ be a continuous mapping. In order that there should exist a continuous mapping $g: X/A\rightarrow Y$ such that $g\circ \rho =f$ , it is necessary and sufficient that $f$ should be constant on A.

If the set A is closed (resp. open) in X, the mapping $\rho$ is closed (resp. open) and induces a homeomorphism of X-A onto its image. If A is a closed and quasi-compact subset of X, the mapping $\rho$ is proper (TG, I, p. 75, th. 1).

If the set A is empty, $\rho$ is a homeomorphism. If it is not empty, A is a point of $X/A$ called the base point of $X/A$ and denoted by $s_{X/A}$ . The space $X/A$ is then identified with the space obtained by attaching the space X to the space $\{s_{X/A}\}$ by means of the unique mapping from A into $\{s_{X/A}\}$ .

Proposition 10 01WH

Let X be a topological space and let A be a subset of $x$ . Suppose that there exists a homotopy $\sigma : X\times \mathbf{I}\rightarrow X$ with origin Id $_X$ , constant on $A\times \{1\}$ and such that $\sigma (A\times \mathbf{I})\subset A$ . Then the canonical mapping $\rho$ of X into $X/A$ is a homotopy equivalence.

Let $f$ be the terminal mapping of $\sigma$ . It is a continuous mapping of X into X, homotopic to Id $_X$ . It is constant on A; there therefore exists a unique continuous mapping $g: X/A\rightarrow X$ such that $g\circ \rho =f$ . On the other hand, since $\sigma (A\times \mathbf{I})\subset A$ , there exists a mapping $\sigma '$ of $X/A\times \mathbf{I}$ into $X/A$ such that $\sigma '(\rho (x), t) =\rho (\sigma (x, t))$ for every $x\in X$ and every $t\in \mathbf{I}$ . By I, p. 19, prop. 10, the mapping $\sigma '$ is continuous. It is a homotopy joining the identity mapping of $X/A$ to the mapping $\rho \circ g$ . Consequently, the mappings $g$ and $\rho$ are homotopy inverses of one another.

Remark 01WI

Let X be a topological space and let A be a subset of X.

1) If the canonical mapping of X onto $X/A$ is a homotopy equivalence, there exists a homotopy $\sigma : X\times \mathbf{I}\rightarrow X$ whose initial mapping is Id $_X$ and which is constant on $A\times \{1\}$ . But it may happen that there exists no such homotopy satisfying moreover the condition $\sigma (A\times \mathbf{I})\subset A$ (III, p. 322, exerc. 4).

2) There may exist a homotopy joining the identity mapping of X to a mapping of X into X which is constant on A, without the spaces X and $X/A$ being homotopy equivalent (III, p. 325, exerc. 14).

3) Suppose that A is contractible. If the pair $(X,A)$ has the homotopy extension property, the canonical mapping of X onto $X/A$ is a homotopy equivalence. Let in fact $\sigma : A\times \mathbf{I}\rightarrow A$ be a homotopy whose initial mapping is the identity mapping of A and whose terminal mapping is a constant mapping. There then exists a homotopy $\sigma '$ with initial mapping Id $_X$ which extends $\sigma$ . The assertion then follows from proposition 10.

4) Let X and Y be topological spaces, let A be a nonempty subspace of X and let B be a subspace of Y. Let $f: X\rightarrow Y$ be a continuous mapping such that $f(A)\subset B$ . Let $\varphi : X/A\rightarrow Y/B$ denote the continuous mapping deduced from $f$ by passing to quotients. If $f$ defines a homotopy equivalence of the pair $(X,A)$ onto the pair $(Y,B)$ , the mapping $\varphi$ is a homotopy equivalence of the pair $(X/A, s_{X/A})$ onto the pair $(Y/B, s_{Y/B})$ . Let P be a topological space reduced to a point, and let $\alpha$ and $\beta$ be the constant mappings of A and B into P. Observe that the mapping $\varphi$ is identified with the mapping of $P\cup_{\alpha}X$ into $P\cup_{\beta}Y$ deduced from $f$ and the identity mapping of P. The assertion then follows from corollary 1 of III, p. 249.

10. Cone of a mapping

Let X and Y be topological spaces and let $f$ be a continuous mapping of X into Y. Let Cyl( $f$ ) be the cylinder of the mapping $f$ . Let $\alpha_f$ denote the canonical mapping of $X\times \mathbf{I}$ into Cyl( $f$ ) and $f_0$ the mapping $x\mapsto \alpha_f(x,0)$ of X into Cyl( $f$ ); these mappings induce homeomorphisms of $X\times \mathbf{I}$ and X respectively onto their images in Cyl( $f$ ).

Definition 11 01WJ

The cone of the mapping $f$ is the topological space deduced from Cyl( $f$ ) by contracting $f_0(X)$ ; it is denoted by Côn $(f)$ .

Let $\beta '_f: Y\rightarrow$ Côn $(f)$ denote the composition of the canonical mapping $\beta_f: Y\rightarrow$ Cyl( $f$ ) and the canonical surjection of Cyl( $f$ ) onto Côn $(f)$ . The mapping $\beta '_f$ is continuous and defines a homeomorphism of Y onto a closed subset of Côn $(f)$ , called the base of the cone, and which we shall thus identify with Y.

Let $\alpha '_f: X\times \mathbf{I}\rightarrow$ Côn $(f)$ denote the composition of the mapping $\alpha_f$ and the canonical surjection of Cyl( $f$ ) onto Côn $(f)$ . The mapping $\alpha '_f$ is a homotopy whose origin is a constant mapping and whose term is the mapping $\beta '_f\circ f$ .

If X is empty, the mapping $\beta '_f$ is a homeomorphism of Y onto Côn $(f)$ .

Suppose that X is not empty. Let $s$ then denote the base point of the space Cyl( $f$ ) $/f_0(X)$ ; it is said to be the vertex of the cone Côn $(f)$ . Since $f_0(X)$ is closed in Cyl( $f$ ), the canonical mapping $\pi :$ Cyl( $f$ ) $\rightarrow$ Côn $(f)$ induces a homeomorphism of Cyl( $f$ ) $-f_0(X)$ onto Côn $(f)-\{s\}($ III, p. 252). The vertex of the cone Côn $(f)$ does not belong to its base; the canonical injection of Y into Côn $(f)-\{s\}$ is a homotopy equivalence (III, p. 239, prop. 6).

Let $\sigma_f:$ Cyl( $f$ ) $\times \mathbf{I}\rightarrow$ Cyl( $f$ ) be the canonical contraction of the cylinder of $f$ onto its base. For $c\in$ Cyl( $f$ ) $-f_0(X)$ and $t\in \mathbf{I}$ , we have $\sigma_f(c, t)\notin f_0(X)$ . Hence there exists a unique mapping $\sigma '_f: ($ Côn $(f)-\{s\})\times \mathbf{I}\rightarrow$ Côn $(f)-\{s\}$ such that $\sigma '_f(\pi (c), t) =\pi (\sigma_f(c, t))$ for $c\in$ Cyl( $f$ ) $-f_0(X)$ and $t\in \mathbf{I}$ . It is continuous and is a strong contraction of Côn $(f)-\{s\}$ onto Y. It is said to be the canonical contraction of Côn $(f)-\{s\}$ onto its base. Its term is a retraction of Côn $(f)-\{s\}$ onto Y, called the canonical retraction of the cone deprived of its vertex onto its base.

Proposition 11 (Universal Property of Cones)

Let Z be a topological space, let $\beta : Y\rightarrow Z$ be a continuous mapping and let $\alpha : X\times \mathbf{I}\rightarrow Z$ be a homotopy whose origin is a constant mapping and whose term is equal to $\beta \circ f$ . There exists a unique continuous mapping $\varphi$ of Côn $(f)$ into Z such that $\alpha =\varphi \circ \alpha '_f$ and $\beta =\varphi \circ \beta '_f$ .

From the universal property of cylinders (III, p. 238, prop. 4), there exists a unique continuous mapping $h:$ Cyl( $f$ ) $\rightarrow$ Z such that $\alpha =h\circ \alpha_f$ and $\beta =h\circ \beta_f$ . Since the source of $\alpha$ is a constant mapping, the restriction of $h$ to the subspace $\alpha_f(X\times \{0\})$ is constant. There therefore exists a unique continuous mapping $\varphi :$ Cone $(f)\rightarrow Z$ such that $h=\varphi \circ \pi$ , where $\pi$ denotes the canonical surjection of Cyl( $f$ ) onto Cone $(f)$ . The mapping $\varphi$ satisfies $\alpha =\varphi \circ \alpha '_f$ and $\beta =\varphi \circ \beta '_f$ and it is the only one having these properties, since the images of $\alpha '_f$ and $\beta '_f$ cover Cone $(f)$ .

Example 1 01WK

Let X be a topological space. One denotes by C(X), and calls the cone of the space X, the cone of the mapping Id $_X$ ; it is the space obtained from $X\times \mathbf{I}$ by contracting $X\times \{0\}$ . Let $\pi$ be the canonical mapping of $X\times \mathbf{I}$ into C(X); the image $C'(X)$ of $X\times [0,1[$ under $\pi$ is called the open cone of the space X.

Assume that X is not empty. Then the cone C(X) is not empty and its base point is again called the vertex of the cone.

The mapping $((x, t), u)\mapsto (x, t(1-u))$ of $(X\times \mathbf{I})\times \mathbf{I}$ into $X\times \mathbf{I}$ is a homotopy joining the identity mapping of $X\times \mathbf{I}$ to the mapping $(x, t)\mapsto (x,0)$ . One deduces from this, by passing to quotient sets, a mapping $\sigma : C(X)\times \mathbf{I}\rightarrow C(X)$ such that $\sigma (\pi (x, t), u) =\pi (x, t(1-u))$ for $x\in X,t, u\in \mathbf{I}$ . The mapping $\sigma$ is continuous by I, p. 19, prop. 10. The mapping $\sigma$ is then a pointed homotopy at $s$ joining the identity mapping of C(X) to the constant mapping with image $\{s\}$ . Consequently, the cone C(X) is contractible to its vertex $s$ . Since $\sigma (C'(X)\times \mathbf{I})$ is contained in $C'(X)$ , the mapping $\sigma$ defines, by passing to subspaces, a pointed homotopy at $s$ joining the identity mapping of $C'(X)$ to the constant mapping with image $\{s\}$ ; the open cone $C'(X)$ is therefore contractible to $s$ .

Remark 1 01WL

Let X be a topological space and let A be a subspace of X; let $i$ denote the canonical injection of A into X. The canonical retraction $r_i$ of the cylinder Cyl( $i$ ) onto its base maps the subspace $\alpha_i(A\times \{0\})$ into A. Let $\rho :$ Cone $(i)\rightarrow X/A$ be the continuous mapping deduced from it by passing to the quotient spaces. Suppose that the pair $(X,A)$ has the homotopy extension property. By III, p. 243, Theorem 1, the mapping $r_i$ defines a homotopy equivalence of the pair (Cyl( $i$ ) $, \alpha_i(A\times \{0\})$ ) with the pair $(X,A)$ . By Remark 4 of III, p. 253, the mapping $\rho$ is then a homotopy equivalence of the pair (Cone $(i), s$ ) with the pair $(X, s_{X/A})$ . In particular it is a homotopy equivalence.

Remark 2 01WM

Let X and Y be topological spaces and let $f: X\rightarrow Y$ be a continuous mapping. The canonical mapping $\alpha '_f: X\times \mathbf{I}\rightarrow$ Cone $(f)$ is constant on $X\times \{0\}$ and therefore defines a continuous mapping $\gamma_f: C(X)\rightarrow$ Cone $(f)$ . The restriction of the mapping $\gamma_f$ to $C'(X)$ is injective and strict, and defines by passing to the subspaces a homeomorphism of the open cone $C'(X)$ onto the complement of Y in Cone $(f)$ .

Let us identify the base of the cone C(X) with the space X and let $u$ denote the continuous mapping of $Y\cup_fC(X)$ into Cone $(f)$ deduced from the mappings $\beta '_f: Y\rightarrow$ Cone $(f)$ and $\gamma_f: C(X)\rightarrow$ Cone $(f)$ . Conversely, let $v:$ Cone $(f)\rightarrow Y\cup_fC(X)$ be the unique continuous mapping such that $v\circ \alpha '_f$ is the canonical mapping of $X\times \mathbf{I}$ onto C(X) and $v\circ \beta '_f$ is the canonical mapping of Y onto $Y\cup_fC(X)$ . The mappings $u$ and $v$ are inverse homeomorphisms of one another.

Exercises

See the exercises for § 1.