Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

I am trying to follow May's proof of the above proposition, but I a do not quite understand one of his conclusions. Namely, he makes states the following during his proof:

By HEP for $j$, there is a homotopy inverse $g'$ for $f$ such that $g'\circ j=i\circ e$.

How does May conclude this based on the fact $j$ is a cofibration? He is hiding a few steps and I just cannot connect the dots, especially since the definition of a cofibration involves always being able to complete a certain type of diagram. I have been trying for a while to find the right diagram, but to no avail. If possible, could someone help illuminate what is happening here?

Here is the whole proof for context: enter image description here

share|cite|improve this question
What is $e$ here? It looks like it should be $e:B \to A$? – Aaron Mazel-Gee Nov 2 '12 at 16:50
@aaronmazel-gee that's right, $e$ is an arbitrary homotopy inverse of $d$. – Holdsworth88 Nov 2 '12 at 17:54
up vote 4 down vote accepted

Actually one can say more: if $d'$ is a homotopy inverse of $d$ and (omitting the $\circ$'s) $h:d' d\simeq 1, k: d d' \simeq 1$, then there is a homotopy inverse $f'$ of $f$ with homotopies $h':f' f \simeq 1, k': f f' \simeq 1$ such that $f'$ extends $d'$, $h'$ extends $h$, while $k'$ extends the composite of the homotopies

$$ d' d = d' d 1_A \simeq d'd d' d \simeq d'1_Bd \simeq 1_A$$ determined by $h,k$.

Comments: 1) This result was arrived at by generalising the classic proof that a homotopy equivalence of spaces induces an isomorphism of homotopy groups. (The point is that the homotopy equivalence is not given as a homotopy equivalence of spaces with base points.) A crucial technique in the proof is the operation of the fundamental groupoid on the higher homotopy groups. When the base point in a sphere is replaced by a cofibration $i: A \to X$ one replaces a fundamental groupoid of $Y$ by the track groupoid $\pi_1 Y^A $ of homotopy classes of homotopies $A \to X$, and let this act on the family of homotopy classes $[(X,i),(Y,u)]$ for all $u:A \to Y$.

2) The advantage of stating the precise control of the homotopies is that it makes it easy to glue homotopy equivalences along the spaces $A,B$. That was how I found the Gluing Lemma for homotopy equivalences for the 1968 edition of Topology and Groupoids, and which is in Section 7.4 of T&G; also it is easy to generalise from gluing two spaces to gluing $n$ spaces (Exercise 1 of Section 7.4).

3) The curious composite of homotopies is related to the elementary lemma in category theory that a right inverse of an isomorphism is itself an isomorphism. This can be applied to homotopy equivalences of spaces instead of isomorphisms, but the relevance comes when you try to identify the homotopies. I expect that the above statement on $k'$ can be recovered from Peter May's proof.

4) I don't think it is possible to get in general the homotopy $k'$ to extend $k$ but I do not have a counterexample. (One has been given in the dual fibration case.)

share|cite|improve this answer

Here's a path I think will work. This material is pretty new to me as well, though, so I don't claim this is the geodesic route to seeing May's point. In any case:

We've got a diagram $$\begin{matrix} A&\stackrel{h}{\longrightarrow}&X^I\\ \downarrow^i&&\downarrow^{p_0}\\ X&\stackrel{\text{id}_X}{\longrightarrow}&X\end{matrix}$$ $h$ is a homotopy between $i$ and $ied$ for $d$ any homotopy inverse of $e$. The diagram commutes since $h(a)(0)=i(a)$ for $a\in A$.

Since $i$ is a cofibration we can extend this homotopy to $\bar{h}:X\to X^I$ so that $p_1 \bar{h}i=p_1 h=ied$. So $p_1 \bar{h}$ sits over $ed$; meanwhile it's homotopic to $\text{id}_X$, and thus homotopic to $gf$ for $g$ any homotopy inverse of $f$. Set $p_1\bar{h}=g'f$.

Then $ied=g'fi=g'jd$, where the first equality holds because $g'$ sits over $ed$ and the second by the commutativity of the original given square. Then $iede=g'jde$ and since $de$ is homotopic to $\text{id}_B$ we can homotopy this over to $ie=g'j$.

share|cite|improve this answer
I'm not too sure about this. For example, $h$ is a map from $A$ to $B^I$. Also, $ied$ cannot equal $g'fi$ since they don't map to the same spaces. – Holdsworth88 Nov 3 '12 at 11:38
$i$ and $ied$ are maps $A\to X$, so $h$ is a homotopy of maps into $X$, i.e. a map to $X^I$-why do you want $B^I$? $d: A\to B, e: B\to A, i: A\to X, f: X\to Y,$ and $g':Y\to X$. I made an error calling $p_1\bar{h} g'$ initially: it should be $g'f,$ but given that $g'fi=ied$ is just the definition of $\bar{h}$. – Kevin Carlson Nov 3 '12 at 19:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.