Who proved that existence of a retraction $r:X\times\mathbb{I}\rightarrow X\times\left\{ 0\right\} \cup A\times\mathbb{I}$ was sufficient for HEP?

Question

It is well known that the existence of a retraction $r:X\times\mathbb{I}\rightarrow X\times\left\{ 0\right\} \cup A\times\mathbb{I}$ is necessary to make $\left(X,A\right)$ a pair having the homotopy extension property (HEP). In Note on cofibrations written in 1966 by Arne Strøm I meet the remark that it is sufficient under the extra condition that $A$ is closed. In Hatcher's Algebraic Topology it is shown that this extra condition can be dropped. Who was the first to find that out, and when did that happen?

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My last comment is not immediately visible on my screen and I do not want to miss this opportunity. So here I repeat it (probably unnecessary) as an addition of my original question. Indeed, as professor Brown remarked, that is what the debate is about: if $X\times\left\{ 0\right\} \cup A\times\mathbb{I}$ is a retract of $X\times\mathbb{I}$ then the pair $\left(X\times\left\{ 0\right\} ,A\times\mathbb{I}\right)$ has the gluing property. Professor Brown, I would like to attend you on a nice consequence of the proof. In Topology and Groupoids (nice book!) 7.2.4 it is shown that $1_{B}\times i$ is a cofibration if $i$ is, but this under the extra condition that $B$ is locally compact. This extra condition can be left out! If $X\times\left\{ 0\right\} \cup A\times\mathbb{I}$ is a retract of $X\times\mathbb{I}$ then $B\times X\times\left\{ 0\right\} \cup B\times A\times\mathbb{I}$ is a retract of $B\times X\times\mathbb{I}$, and that is enough.

Answer 1

This result is proved in a second paper by Arne Strøm, "Note on Cofibrations II" in Math. Scand. 22 (1968), 130-142. As far as I am aware this is the original source, and in any case it was the source for the proof given in the appendix of my book (added in 2009 to the online version of the book). I will add this paper to the list of references in the book -- I don't know why I didn't do this before.

Answer 2

The following result is part of 7.2.4 of my book "Topology: a geometric account of general topology, homotopy types and the fundamental groupoid" Ellis Horwood, (1988) (which itself was a revised version of "Elements of Modern Topology" (McGraw Hill, 1968)).

Let $: A \to X$ be a map of spaces. Let $M(i)$ be the pushout of the two maps $i \times 1: A \times \{0\} \to X \times \{0\}$, $A \times \{0\} \to A \times I$ where $I=[0,1]$. This pushout determines a map $\mu: M(i) \to X \times I$. Then $i$ is a cofibration if and only if there is a map $\rho: X \times I \to M(i)$ such that $\rho \mu =1$.

A revised version of that book is available as Topology and Groupoids.

But I first studied these ideas from papers of J.H.C. Whitehead and others in giving MSc lectures on Homotopy Theory in 1960 at Liverpool.

@Stefan H.:Further comments (September 15): I am puzzled by the comments following my answer, but perhaps am missing something. To put the matter in context, let us say a pair $(C,D)$ of subspaces of a space $Y$ satisfy the gluing property if the following diagram of inclusions $$\matrix{C \cap D& \to & D \cr \downarrow & & \downarrow \cr C &\to & C \cup D}$$ is a pushout of spaces. This is true for example, 2.5.11 of "Topology and Groupoids", if $$C \backslash D \subseteq Int(C), D \backslash C \subseteq Int(D).$$

Now let $i: A \to X$ be an inclusion of spaces. My question is whether the pair $(A \times I, X \times \{0\})$ satisfies always satisfies the gluing property. If so, then giving maps $A \times I \to Y, X \times \{0\} \to Y$ is equivalent to giving a map $A \times I \cup X \times \{0\} \to Y$. In terms of the notation above, this is equivalent to saying that the canonical map $M(i) \to A \times I \cup X \times \{0\} $ is always a homeomorphism, as it is when $A$ is closed in $X$. However $$(X \times \{0\}) \backslash (A \times I)= (X \backslash A) \times \{0\},$$ so the above condition for gluing does not necessarily hold.

I am unable to prove that the gluing condition holds if $A \times I \cup X \times \{0\}$ is a retract of $X \times I$. I think this is what the debate is about.

Later: I am just about to study Hatcher's correction in his Appendix.

Actually he does prove in this new Appendix that if the retraction exists then the pair $(A \times I, X \times 0)$ satisfies the gluing property, and so $(X,A)$ has the HEP! The proof is delicate, and I expect is due to Alan Hatcher. Good to have this sorted. Are there circumstances where the existence of this retraction, rather than to $M(i)$, is the primary information?