The topological space $\mathbf{I}$ being locally compact, the proposition follows from TG, X, p. 28, th. 3.

We denote $o$ and call origin mapping the mapping $c\mapsto c(0)$ of Λ(X) into X; we denote $e$ and call term mapping the mapping $c\mapsto c(1)$ of Λ(X) into X. The mappings $o$ and $e$ are continuous (TG, X, p. 27, remark 1). The pairs of juxtaposed paths in X are the elements of the fibered product of the X-spaces $(\Lambda (X), e)$ and $(\Lambda (X), o)$.

The mapping $t\mapsto 1-t$ is a homeomorphism of the space $\mathbf{I}$ onto itself; the first assertion follows.

Let C denote the fibered product $(\Lambda (X), e)\times_X(\Lambda (X), o)$. Let $\gamma : C\rightarrow$ Λ(X) be the mapping $(c, d)\mapsto c*d$ and $\delta : C\times \mathbf{I}\rightarrow$ X the mapping $((c, d), t)\mapsto (c*d)(t)$. The restrictions of the mapping $\delta$ to $C\times [0,\frac{1}{2}]$ and $C\times [\frac{1}{2},1]$ are continuous by virtue of formula (1) and of prop. 1. The mapping $\delta$ is therefore continuous (TG, I, p. 19, prop. 4), as is the mapping $\gamma$ (prop. 1). We have $c(t) = (c*d)(\frac{t}{2})$ and $d(t) = (c*d)(\frac{1+t}{2})$, hence $\gamma$ is injective.

Let $g$ be a path in X; for $t\in \mathbf{I}$, put

$c_g(t) =g(\frac{t}{2})$ and $d_g(t) =g(\frac{1 + t}{2})$.

The mappings $c_g$ and $d_g$ thus defined are juxtaposed paths in X and we have $c_g*d_g=g$. Moreover, the mappings $g\mapsto c_g$ and $g\mapsto d_g$ are continuous mappings of the space Λ(X) into itself (prop. 1). It follows that the mapping $\gamma$ is a homeomorphism.

Let X be a path-connected space, $f: X\rightarrow Y$ a continuous mapping. Let $x$ and $y$ be two points of $f(X)$; let $x'$ and $y'$ be points of X such that $f(x') =x$ and $f(y') =y$. There exists a path $c$ in X whose origin is $x'$ and whose term is $y'$. Then, $f\circ c$ is a path in $f(X)$ with origin $x$ and term $y$.

Since every interval of $\mathbf{R}$ is connected, the image of a path is connected (TG, I, p. 82, prop. 4). The empty space is connected. Since a nonempty path-connected space is the union of the images of the paths issuing from one of its points, it is connected (TG, I, p. 81, prop. 2).

Taking into account the identification (III, p. 256, remark 1) of the paths in X with the homotopies between mappings of a topological space P reduced to a point into X, Proposition 1 of III, p. 230 gives:

The path-connected components of a topological space X are called the equivalence classes for the above relation. A path-connected component is a path-connected space. Let $x$ be a point of X. The path-connected component of $x$ is the union of the path-connected subspaces of X containing $x$. It is also the union of the images of the paths with origin $x$ in X.

$(\pi_0$(pr$_j$))$:\pi_0(\prod_{j\in J}Y_j)\rightarrow \prod_{j\in J}\pi_0(Y_j)$

is bijective. In particular, the product space of a family of path-connected spaces is path-connected.

This follows from Proposition 3 of III, p. 231 where one takes for X a space reduced to a point.

The first assertion follows from prop. 4. Let us prove the second assertion. Let X be a locally path-connected topological space and let C be a path-connected component of X. Every point of C possesses a path-connected neighbourhood, hence contained in C; consequently, C is an open subset of X. The path-connected components forming a partition of X, such a component is also closed. Since it is connected (III, p. 258, prop. 4), it is a connected component (TG, I, p. 83). The third assertion follows from the second.

Thus we see that there is no ambiguity in saying that a topological space is connected and locally path-connected.

The first assertion follows immediately from def. 4 and the definition of an étale mapping (I, p. 28, def. 2). The second follows from it by prop. 7.

In order that a topological space X be locally path-connected, it is necessary and sufficient that every path-connected component of an open set of X be an open set of X (cf. I, p. 85, proof of prop. 11). If the space X is locally path-connected, every point of X therefore possesses a fundamental system of open path-connected neighbourhoods.

Let X be a locally path-connected space, let R be an equivalence relation in X and let $\varphi : X\rightarrow X/R$ be the canonical mapping. It suffices to prove that the path-connected components of an open subset of $X/R$ are open. Let thus A be an open subset of $X/R$ and let C be a path-connected component of A. Let $x\in \overset{-1}{\varphi}(C)$, and let K be the path-connected component of $x$ in $\overset{-1}{\varphi}(A)$. The set $\varphi (K)$ is path-connected (III, p. 258, prop. 3), contained in A and contains $\varphi (x)$; hence $\varphi (K)\subset C$, whence $K\subset \overset{-1}{\varphi}(C)$. This proves that $\overset{-1}{\varphi}(C)$ is the union of path-connected components of $\overset{-1}{\varphi}(A)$. Since X is locally path-connected and $\overset{-1}{\varphi}(A)$ is open in X$,\overset{-1}{\varphi}(C)$ is open in X. Consequently, C is open in $X/R$. This proves the proposition.

Let $(X_j)_{j\in J}$ be a family of locally path-connected spaces which are connected, with the exception of a finite number of them. Let $x=$ $(x_j)_{j\in J}$ be a point of the product space X = $\prod_{j\in J}X_j$. By assumption, a fundamental system of neighbourhoods of $x$ in X consists of sets of the form $V =\prod_{j\in J}V_j$ where, for every $j\in J$, $V_j$ is a path-connected neighbourhood of $x_j$ in $X_j$ and where $V_j= X_j$ except for a finite set of indices $j\in J$ (TG, I, p. 24). These sets being path-connected (III, p. 260, prop. 6), X is locally path-connected.

Every point of $\mathbf{R}$ possesses a basis of neighbourhoods formed by intervals; consequently, $\mathbf{R}$ is locally path-connected. According to Proposition 9, the numerical space $\mathbf{R}^n$ is locally path-connected, for every integer $n\geqslant 1$. Moreover, an open subset of a locally path-connected topological space is again path-connected, whence the corollary.

We already know that the origin mappings $o$ and term $e$ of Λ(X) into X are continuous (III, p. 257), whence the first assertion.

$c'(t)\in V_0$ for $0\leqslant t\leqslant \varepsilon$,

(2) $c'(t) =c(t)$ for $\varepsilon \leqslant t\leqslant 1-\varepsilon$,

$c'(t)\in V_1$ for $1-\varepsilon \leqslant t\leqslant 1$,

belongs to W.

By definition of the topology of compact convergence (TG, X, p. 26, déf. 1), there exists a finite set J, a family $(U_j)_{j\in J}$ of open sets in X and a family $(K_j)_{j\in J}$ of compact subsets of $\mathbf{I}$ such that the set $W'$ of paths $c'$ satisfying $c'(K_j)\subset U_j$ for every index $j$ is a neighbourhood of $c$ contained in W. Let us then denote by $A_0$ (resp. $A_1$) the set of indices $j$ such that $0\in K_j$ (resp. $1\in K_j$); put $V_0=\bigcap_{j\in A_0}U_j$ and $V_1=\bigcap_{j\in A_1}U_j$.

Since the mapping $c$ is continuous, there exists a real number $\varepsilon \in$ $]0,1/2]$ such that $c([0, \varepsilon ])\subset V_0,c([1-\varepsilon ,1])\subset V_1,[0, \varepsilon ]\cap K_j=\emptyset$ for all $j\notin A_0$ and $[1-\varepsilon ,1]\cap K_j=\emptyset$ for all $j\notin A_1$. Let then $c'$ be a path satisfying the conditions (2). Let us prove that $c'\in W'$. Let $j\in J$ and let $t\in K_j$. If $\varepsilon \leqslant t\leqslant 1-\varepsilon ,c'(t) =c(t)$ belongs to $U_j$. If $0\leqslant t\leqslant \varepsilon ,c'(t)\in V_0$; by the choice of $\varepsilon$, we have $j\in A_0$, hence $c'(t)\in U_j$. Analogously, if $1-\varepsilon \leqslant t\leqslant 1$, we have $j\in A_1$ and $c'(t)\in V_1\subset U_j$. Thus, $c'(K_j)\subset U_j$ and $c'$ belongs to $W'$, hence to W.

Let us now prove the second assertion of Proposition 10. Suppose that the space X is locally connected by arcs. Let $c$ be a path in X and let W be a neighbourhood of $c$ in $\mathscr{C}_c(\mathbf{I}; X)$. Let $\varepsilon ,V_0$ and $V_1$ be as in the lemma. Let $T_0$ and $T_1$ be connected by arcs neighbourhoods of $c(0)$ and $c(1)$ contained in $V_0$ and $V_1$ respectively. There exists a real number $\theta$ such that $0< \theta < \varepsilon$ and such that $c([0, \theta ])\subset T_0,c([1-\theta ,1])\subset T_1$. Let $x_0\in T_0$ and $x_1\in T_1$; let $c_0$ be a path of origin $x_0$ and of term $c(\theta )$ in $T_0$ and let $c_1$ be a path of origin $x_1$ and of term $c(1-\theta )$ in $T_1$. Put

$c_0(t/\theta )$ for $0\leqslant t\leqslant \theta$,

$c'(t) =c(t)$ for $\theta \leqslant t\leqslant 1-\theta$,

$c_1((1-t)/\theta )$ for $1-\theta \leqslant t\leqslant 1$.

One thus defines a path $c'$ joining $x_0$ to $x_1$ and satisfying the conditions (2). This proves that the image of W in $X\times X$ by the mapping $(o, e)$ contains the neighbourhood $T_0\times T_1$ of $(c(0), c(1))$, whence the proposition.

According to the proposition, the mapping $\varphi : \Lambda (X)\rightarrow X\times X$ defined by $\varphi (c) = (c(0), c(1))$ is open and one has $\Lambda_x(X) =\overset{-1}{\varphi}(\{x\} \times X)$. Consequently, the mapping $\Lambda_x(X)\rightarrow \{x\} \times X$ deduced from $\varphi$ is open (TG, I, p. 30, prop. 2), as is its composite with the second projection pr$_2$.

Let X be a topological space. Call a train in X any nonempty finite sequence $T = (W_i)_{1\leqslant i\leqslant n}$ of connected open subsets of X such that $W_i\cap W_{i+1}=\not\emptyset$ for $1\leqslant i\leqslant n-1$. We say that $n$ is the length of the train T and that the $W_i$ are its wagons. If X is endowed with a distance compatible with its topology, we call the width of the train T the maximum of the diameters of its wagons. We say that the train joins a point $a$ to a point $b$ if $a$ belongs to the first and $b$ to the last wagon. We call a refinement of T any pair $(T', f)$ formed by a train $T'= (W'_j)_{1\leqslant j\leqslant m}$ and a strictly increasing mapping $f:\{0,1, . . . , n\} \rightarrow \{0,1, . . . , m\}$ such that $f(0) = 0,f(n) =m$ and $W'_j\subset W_i$ for $1\leqslant i\leqslant n$ and $f(i-1)< j\leqslant f(i)$.

More precisely, every train T joining $a$ to $b$ possesses a refinement $(T', f)$, where $T'$ is a train of width $\leqslant \varepsilon$ joining $a$ to $b$.

The relation “there exists a train of width $\leqslant \varepsilon$ joining $x$ to $y$” is an equivalence relation between $x$ and $y$ in X. The equivalence class of a point $x$ contains every connected open neighbourhood of $x$ of diameter $\leqslant \varepsilon$, and $x$ possesses such a neighbourhood since X is locally connected. The equivalence classes for this relation are therefore open, and consequently also closed. There is at most one of them, since X is connected. This proves the first assertion.

Let $T = (W_i)_{1\leqslant i\leqslant n}$ be a train in X joining $a$ to $b$. Put $x_0=a$, $x_n=b$ and choose for $1\leqslant i\leqslant n-1$ a point $x_i$ in the nonempty set $W_i\cap W_{i+1}$. For $1\leqslant i\leqslant n$, the open set $W_i$ is connected and locally connected and $x_{i-1},x_i$ are two of its points; there exists by the preceding paragraph a train $(W_{i,k})_{1\leqslant k\leqslant m_i}$ in $W_i$, of width $\leqslant \varepsilon$, joining $x_{i-1}$ to $x_i$.

Put $m=m_1+\cdots +m_n$. For $1\leqslant j\leqslant m$, put $W'_j= W_{i,k}$, where $(i, k)$ is the unique pair of integers such that $1\leqslant i\leqslant n,1\leqslant k\leqslant m_i$ and $j=m_1+\cdots +m_{i-1}+k$. For $0\leqslant i\leqslant n$, put $f(i) =m_1+\cdots +m_i$. Then $T'= (W'_j)_{1\leqslant j\leqslant m}$ is a train of width $\leqslant \varepsilon$ in X joining $a$ to $b$, and $(T', f)$ is a refinement of T.

Let us now prove Proposition 11. Equip the connected and locally connected space X with a distance $d$, compatible with its topology, for which it is complete. Let $a$ and $b$ be points of X. Lemma 1 permits us to construct by induction sequences $(T_s)_{s\geqslant 0}$ and $(f_s)_{s\geqslant 1}$ such that, for every $s\geqslant 0$, $T_s= (W_{s,i})_{1\leqslant i\leqslant n_s}$ is a train of width $\leqslant 2^{-s}$ joining $a$ to $b$ and $(T_{s+1}, f_{s+1})$ is a refinement of $T_s$.

We can choose by induction, for $s\geqslant 0$, a strictly increasing mapping $g_s:\{0,1, . . . , n_s\} \rightarrow \mathbf{I}$ such that $g_s(0) = 0$ and $g_s(n_s) = 1$, in such a way that $g_{s+1}\circ f_s=g_s$. Define, for $s\geqslant 0$, a subset $A_s$ of $\mathbf{I}\times X$ by putting

The sequence $(A_s)_{s\geqslant 0}$ is decreasing: indeed, for every integer $j\in \{1, . . . , n_{s+1}\}$, there exists a unique integer $i\in \{1, . . . , n_s\}$ such that $f_s(i-1)< j\leqslant f_s(i)$, and one has

Let $t\in \mathbf{I}$. For every $s\geqslant 0$, denote by $A_s(t)$ the set of the $x\in X$ such that $(t, x)\in A_s$. The set $A_s(t)$ is either one of the cars, or the union of two consecutive cars of the train $T_s$; it is therefore a nonempty subset of X, of diameter $\leqslant 2^{1-s}$. The sequence $(A_s(t))_{s\geqslant 0}$ is decreasing. The set of its terms is a Cauchy filter basis. This one converges to a point $c(t)$ since the metric space X is complete.

Since $a$ belongs to each of the sets $A_s(0) = W_{s,1}$, we have $c(0) =a$; analogously $c(1) =b$. Let $t\in \mathbf{I}$ and let $s$ be an integer $\geqslant 0$. The point $t$ possesses in $\mathbf{I}$ a neighbourhood V of one of the following forms: $[g_s(0), g_s(1)[$, $]g_s(i-1), g_s(i+ 1)[$ for an integer $i$ such that $1\leqslant i\leqslant$ $n_s-1$, or $]g_s(n_s-1), g_s(n_s)]$. According to the case, the set $c(V)$ is contained in the closure of the first car, of the union of two consecutive cars, or of the last car of the train $T_s$. It is therefore of diameter $\leqslant 2^{1-s}$, and we have $d(c(t), c(t'))\leqslant 2^{1-s}$ for $t'\in V$. This proves that the mapping $c:\mathbf{I}\rightarrow$ X is continuous. Thus $c$ is a path in X joining $a$ to $b$, and X is arcwise connected.

Let X be such a space and let U be an open and connected subset of X. According to Lemma 2 below, there exists a distance on U compatible with its topology for which U is complete. Since U is locally connected, it follows from Proposition 11 that U is arcwise connected.

We shall take up again the arguments of the proof of prop. 2 of TG, IX, p. 57. We may suppose U distinct from X. Let V be the subset of the product $\mathbf{R}\times X$ formed by the points $(t, x)$ such that $t d(x,X-U) = 1$ ; the subspace V of $\mathbf{R}\times X$ is closed and the mapping $(t, x)\mapsto x$ from V into U is a homeomorphism (TG, IX, p. 13, prop. 3). There exists on $\mathbf{R}\times X$ a distance $d'$ compatible with its topology for which $\mathbf{R}\times X$ is complete (TG, IX, p. 15, cor. 2 and TG, II, p. 17, prop. 10). The space V is complete for the distance induced by $d'$ (TG, II, p. 16, prop. 8), whence the lemma.

It suffices to prove the first assertion (III, p. 260, prop. 7). Let X be a locally compact and locally connected metric space. The open, connected and relatively compact subsets of X constitute a basis of the topology of X. Let U be such a set; since U is an open subset of its closure, which is a compact metric space, hence complete, there exists according to Lemma 2 a distance on U compatible with its topology for which U is complete. It follows from prop. 11 of III, p. 264 that U is arcwise connected, whence the corollary.

According to (TG, IX, p. 18), it suffices to prove that, for every point $x$ of X and every sequence $(x_n)_{n\geqslant 1}$ of points of X which tends to $x$, the sequence $(f(x_n))_{n\geqslant 1}$ tends to $f(x)$. By deleting, if necessary, the first terms of the sequence $(x_n)_{n\geqslant 1}$, we reduce to the case where all the terms of the sequence belong to the same arc-connected neighbourhood of $x$ in X. According to the lemma below, there then exists a path $c:\mathbf{I}\rightarrow X$ such that $c(0) =x$ and $c(1/n) =x_n$ for $n\geqslant 1$. If the mapping $f$ is continuous by arcs, the mapping $f\circ c$ is continuous and the element $f(x) =f(c(0))$ is the limit of the sequence $(f(c(1/n)))_{n\geqslant 1}$, that is to say of the sequence $(f(x_n))_{n\geqslant 1}$, whence the corollary.

Let $(W_m)_{m\geqslant 1}$ be a fundamental system of neighbourhoods of $x$. Put $V_0= X$ and for every $m\geqslant 1$, let $V_m$ be an arc-connected neighbourhood of $x$ contained in $V_{m-1}\cap W_m$.

For every integer $n\geqslant 1$, denote by $m_n$ the largest integer $m\leqslant n$ such that $x_k\in V_m$ for every $k\geqslant n$. The sequence $(m_n)_{n\geqslant 1}$ is increasing by definition; it tends to infinity, because the sequence $(x_n)_{n\geqslant 1}$ tends to $x$. For every integer $n\geqslant 1$, let $c_n:\mathbf{I}\rightarrow V_{m_n}$ be a path with origin $x_{n+1}$ and endpoint $x_n$ in $V_{m_n}$. Define a mapping $c:\mathbf{I}\rightarrow X$ by putting $c(0) =x$ and $c(t) =c_n(n(n+ 1)t-n)$ if $1/(n+ 1)< t\leqslant$ $1/n$. We have $c(1/n) =x_n$ for every $n\geqslant 1$. The mapping $c$ is therefore continuous on every interval of the form $[1/(n+ 1),1/n]$ with $n\geqslant 1$, hence on the interval $]0,1]$. If $t\leqslant 1/n$, the point $c(t)$ belongs to $V_{m_n}$; the mapping $c$ is therefore continuous at 0.

Let $b$ be a point of B; let us prove that $s$ is continuous at the point $b$. Since $p$ is an étale mapping, there exist a neighbourhood V of $b$ and a continuous local section $s'$ of $p$ defined in V such that $s'(b) =s(b)$ (I, p. 33, prop. 9). Since B is locally path-connected, we may suppose that V is path-connected. For every path $c$ in V, with origin $b$, the mappings $s\circ c$ and $s'\circ c$ are two continuous liftings to B of the mapping $c:\mathbf{I}\rightarrow X$ and we have $s\circ c(0) =s'\circ c(0) =s(b)$. By Corollary 1 of I, p. 34, we have $s\circ c=s'\circ c$ and in particular $s\circ c(1) =s'\circ c(1)$. Since every point of V is the end point of a path in V with origin $b$, the mappings $s$ and $s'$ coincide in V. The mapping $s$ is therefore continuous in V.

The mapping pr$_1: E'\times_BE\rightarrow E'$ is étale and separated (I, p. 31, prop. 8 and I, p. 27, prop. 4) and the mapping $x\mapsto (x, f(x))$ is a path-continuous section of it. By Proposition 13, it is continuous, hence $f$ is continuous.

Let X be such a topological space. Let us endow it with a distance compatible with its topology. Since the space X is compact, it is complete for this distance (TG, II, p. 27, th. 1).

Every point of X admits an open and closed neighbourhood of diameter $\leqslant \varepsilon$ (TG, II, p. 32, corollary of prop. 6). Since the space X is compact, it has a finite covering by such sets. Choose one, $(U_i)_{1\leqslant i\leqslant m}$, for which $m$ is minimal. We have $m\geqslant 1$ since X is nonempty. For $1\leqslant i\leqslant m$, denote by $V_i$ the intersection of $U_i$ and the $X-U_k$ for $k < i$. Then $(V_i)_{1\leqslant i\leqslant m}$ is a partition of X, formed of $m$ nonempty open and closed sets of diameter $\leqslant \varepsilon$.

Since X has no isolated point, every nonempty open and closed subset V of X contains at least two points. Since moreover X is compact and totally disconnected, V is the union of two disjoint nonempty open and closed subsets (loc. cit.). It follows, by induction, that for every integer $n\geqslant m$, X admits a partition formed of $n$ nonempty open and closed sets of diameter $\leqslant \varepsilon$.

Let us now finish the proof of the prop. 14. Every nonempty open and closed subspace of X is a compact, totally disconnected metric space and has no isolated point. Lemma 3 therefore allows us to construct by induction a sequence $(J_n)_{n\geqslant 0}$ of finite sets and, for every $n\geqslant 0$, a mapping $\varphi_n$ from the set $C_n= J_0\times \cdots \times J_n$ into the set of nonempty open and closed subsets of X of diameter $\leqslant 2^{-n}$, in such a way that :

(i) For every $n\geqslant 0$, there exists an integer $m_n\geqslant 1$ such that Card(J$_n$) $=$ $2^{m_n}$;

(ii) The family $(\varphi_0(c))_{c\in C_0}$ is a partition of X ;

(iii) For every $n\geqslant 0$ and every $c\in C_n$, the family $(\varphi_{n+1}(c, j))_{j\in J_{n+1}}$ is a partition of $\varphi_n(c)$.

Let us denote by $p_n$ the canonical projection of $C_{n+1}$ onto $C_n$. The sequence C = $(C_n, p_n)_{n\geqslant 0}$ is a sieve (TG, IX, p. 63, def. 8). The topological space associated with this sieve (TG, IX, p. 63) is identified with the topological space J, product of the discrete topological spaces $J_n$. The sieve C and the sequence of mappings $(\varphi_n)_{n\geqslant 0}$ define a strict sieving of the metric space X (TG, IX, p. 63 and p. 64). The mapping $f: J\rightarrow X$ deduced from this sieving is continuous and bijective (TG, IX, p. 65). Since the topological space J is compact (TG, I, p. 63, th. 3) and X is separated, $f$ is a homeomorphism (TG, I, p. 63, cor. 2). Since $J_n$ is homeomorphic to $\{0,1\}^{m_n}$ for every $n\geqslant 0$, J is homeomorphic to $\{0,1\}^{\mathbf{N}}$ (TG, I, p. 25, prop. 2).