A pair $(X,A)$ of topological spaces is said to have the homotopic extension property if $\forall Y$ topological space, $f:X-> Y$ continuous map and $H:A\times[0,1]\rightarrow Y$ homotopy of $f$, there exists $\tilde H : X\times[0,1]\rightarrow Y$ extending H.
In class we gave the following lemma:
$(X,A)$ has the h.e.p. $\Leftrightarrow X\times[0,1]$ retracts onto $(X\times\{0\})\ \cup\ (A\times[0,1])$
I think I need $A$ to be closed for this to work but I am probably wrong, so I'll sketch my argument for "$\Leftarrow$" so maybe someone can spot my mistake.
If $r$ is said retraction, in class we simply defined $\tilde H$ to be $f\circ r$ for on $r^{-1}(X\times \{0\})$ and $H\circ f$ on $r^{-1}(A\times[0,1])$ and then claimed that this is well defined. However, I can't see how this function would be continuous if not by using the gluing lemma, which needs A to be closed in X.
Am I missing something?
Thank you.
The proof for $A$ not necessarily closed is more complicated. See pages 14-15 and Proposition A.18 in Hatcher, which is due to Strøm (Theorem 2 and Lemma 3).