Suppose we have a map $f:X\to Y$ and we form the mapping cylinder $M_f$. Hatcher claims that it is obvious that the pair $(M_f, X \cup Y)$ satisfies the homotopy extension property. Equivalently we could find a retraction of $M_f \times I$ to $M_f\times \{0\} \cup (X \cup Y)\times I$. I don't see how we can get this latter result, however.

  • 4
    $\begingroup$ Have you looked at the proof of the HEP for subcomplexes of a CW-complex? You can adapt the proof directly, and write the extension explicitly. IMO math.stackexchange.com is a more appropriate forum for this kind of question. $\endgroup$ Sep 3, 2011 at 0:16
  • $\begingroup$ This result is in the basis of all homotopy theory, once you understand this you will be able to go directly to the abstract/algebraic homotopy literature ;-) $\endgroup$ Sep 3, 2011 at 1:20
  • $\begingroup$ I can not edit to insert \} in the code. It is too short. $\endgroup$
    – Sigur
    Jul 30, 2015 at 8:14

3 Answers 3


Neil has given an explicit retraction. But it may be useful to note that you can obtain results like this from a combination of some "easier" facts:

  • The pair $(I,\{0,1\})$ has the HEP.

  • If $(L,K)$ has the HEP where $K$ and $L$ are locally compact Hausdorf, and if $Z$ is any space, then $(Z\times L, Z\times K)$ has the HEP.

  • If $(U,A)$ has the HEP, and $g:A\to B$ is any map, then $(V,B)$ has the HEP, where $V$ is the pushout of $U$ along $g$.

(I'll leave the proofs of these as an exercise; you only need the second one for $(L,K)=(I,\{0,1\})$ anyway.) Then note that $M_f$ can be obtained from $X\amalg Y$ by gluing it to a copy of $X\times I$ along $X\times \{0,1\}$.

  • $\begingroup$ I always prefer proofs like this because they remind you that the thing you're trying to prove just boils down to some fact about the unit interval, or something similarly elementary. $\endgroup$ Sep 3, 2011 at 20:20
  • $\begingroup$ Thank you Charles! This really helped me understand after proving these. In your second bullet point, though, is locally compact Hausdorff necessary? Just take $id \times r$ where is $r$ is the assumed retraction. Hatcher also says any space $Z$ will work and he mentions no assumptions on $(L,K)$ except that $K$ will be closed under Hausdorff conditions. Thanks again. $\endgroup$
    – Kyle
    Sep 4, 2011 at 20:00
  • 1
    $\begingroup$ Kyle: it may be the condition I gave is too strong (or maybe it is not really the right condition at all). Here is what I am worried about with your proof: you want to know that the product of $Z$ with $L\times 0\cup K\times I$ is homeomorphic to $(Z\times L\times 0)\cup (Z\times K\times I)$. In general, taking products with a fixed space does not commute with colimits in Top. $\endgroup$ Sep 5, 2011 at 15:38
  • $\begingroup$ I'm not sure this is an issue here. Since the domain of $id\times r : Z\times (L\times I) \to Z\times (L\times I)$ is certainly continuous being the product of continuous maps and our topology being product topology. Then we can just compute that the image is what we want and we get this is actually a retraction. I'll have to think about if those spaces are homeomorphic or not. At a quick glance it would seem you're trying to get at that $Z\times L \times I$ is homeomorphic to $Z\times (L\times I)$ and then the subspaces were looking at are the same sets. Thanks again for commenting :) $\endgroup$
    – Kyle
    Sep 8, 2011 at 4:37
  • $\begingroup$ err sorry i meant to say since $id\times r$ is certainly continous $\endgroup$
    – Kyle
    Sep 8, 2011 at 4:38

I'll assume you want the convention where $M_f$ is $(X\times I)\cup Y$ with $(x,0)$ attached to $f(x)$. Now $M_f\times I=(X\times I^2)\cup(Y\times I)$ with $(x,0,t)$ attached to $(f(x),t)$. We want to retract this onto the space $$ Q=(M_f\times\{0\})\cup(((X\times\{1\})\cup Y)\times I) $$ Note that $X\times\{0\}\times I$ gets identified with part of $Y\times I$ and so is contained in $Q$. Thus $Q=(X\times U)\cup(Y\times I)$, where
$$ U=(\{0,1\}\times I)\cup (I\times\{0\}), $$ and again $(x,0,t)$ is attached to $(f(x),t)$. Now let $r$ be a retraction from $I\times I$ onto $U$, say by radial projection from the point $(1/2,1)$. We can then fit $1\times r:X\times I^2\to X\times U$ together with the identity map on $Y\times I$ to get the required retraction of $M_f\times I$ onto $Q$.


This question is answered by Chapter 7 "Cofibrations", Example 2 on p. 280 of my book `Topology and groupoids' with full proof. In fact it was in the first (1968) edition of this book, published by McGraw Hill.

Other things in that Chapter are a gluing theorem for homotopy equivalences, the exact sequence of a fibration of groupoids, ....

In other Chapters you will find the Phragmen-Brouwer property, the Jordan Curve Theorem, covering morphisms of groupoids, the fundamental groupoid of an orbit space, ...

See http://groupoids.org.uk/topgpds.html


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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