Hatcher's algebraic topology (p. 14) defines that a pair $(X, A)$ where $X$ is a topological space and $A$ an arbitrary subset has the homotopy extension property (HEP) if:
Given a continuous map $f : X \to Z$ to some topological space $X$ and a homotopy $\bar{H}: A \times I \to Z$ of $f|_A$ on $A$ there exists a homotopy $H: X \times I \to Z$ of $f$ such that $H|_{I \times A} = \bar{H}$
(We use the notation $I = [0, 1]$) The book then proceeds to given an equivalent characterization of this property as:
There exists a retraction $r: X \times I \to X \times \{0\} \cup A \times I$
Let's call this condition (R). It's easy to prove (HEP) $\Rightarrow$ (R) and (R) $\Rightarrow$ (HEP) to be pretty easy, too: $(x, 0) \mapsto f(x)$ and $\bar{H}$ as in (HEP) agree on $X \times \{0\} \cap A \times I = A \times \{0\}$, so they glue to a map $F: X \times \{0\} \cup A \times I \to Z$ which we can compose with the retract $r$ from (R) to obtain the desired extension $H: X \times I \to Z$.
However, there is one caveat: The glue map $F$ need not be continuous.
For example, for $X = [0, 1]$, $A = [0, 1)$ and $Z = \mathbb{R}$ we can choose $f$ to be constant $0$ on $[0, 1]$ and $H : [0, 1) \times [0, 1] \to \mathbb{R}$ to be given by $H(x, t) = t/x$. Obviously $H$ can not be extended to $[0, 1]^2$. Note however, that this pair $(X, A)$ does not admit a retraction $r: X \to A$.
Questions: Is the claim (R) $\Rightarrow$ (HEP) in Hatcher correct? That is, can we somehow use the existence of the retraction $r$ to conclude that the glued map $F$ is continuous, or at least that $F \circ r$ is continuous?
If the answer is negative, what are some conditions we can impose on $X$ or $A$ to ensure the equivalence?
What I've got so far:
- Obviously, if $A$ is closed the continuity of $F$ is guaranteed.
- If $X$ is compact, the existence of the retraction $r$ together with the compactness of $X \times I$ implies that $X \times \{0\} \cup A \times I$ and hence also $A$ are closed.
- If $X$ is Hausdorff, (R) also implies that $A$ is closed: Suppose $x \in \overline{A} \setminus A$. Then $r(x) \neq x$ and since $X \times I$ is Hausdorff, we find disjoint open subsets $U, V \subset X \times I$ such that $x \in U$ and $r(x) \in V$. Since $x$ is in the closure of $A$, the open set $U \cap r^{-1}(V)$ contains a point of $A$ which is then not mapped to itself by $r$, contradicting that $r$ is a retraction.
Your doubts are justified. For a cofibred pair $(X,A)$ - which means that the pair has the HEP - there trvially exists a retraction $r: X \times I \to X \times \{0\} \cup A \times I$. What about the converse? Hatcher's argument is too simple - it certainly applies if $A$ is closed, but it has a giant gap for non-closed $A$. Perhaps surprisingly the converse is true for any $A$, but the proof requires much more effort. See