# Homotopy Extension Property involving mapping cylinder [closed]

Suppose we have a map $f:X\to Y~$, $X$ and $Y~$ CW complexes, but $f$ not necessarily a cellular map, 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.

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A good comment from Prof Budney on MO: mathoverflow.net/questions/74406/… –  Dylan Moreland Sep 3 '11 at 1:54
Since Neil Strickland answered the crossposted question on MO, I voted to close the question here. Please don't post on two fora simultaneously. Choose one among the two and post a first time. If you didn't get a satisfactory answer after a few days you can still ask the question on the other forum. If you do so, please say that you asked the question on another platform in order to avoid duplication of efforts. –  t.b. Sep 3 '11 at 9:01