Proposition 7

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

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

and

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

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

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.