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Mar 12, 2014 at 12:34 vote accept archipelago
Jan 31, 2014 at 11:08 history edited Ricardo Andrade CC BY-SA 3.0
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Jan 31, 2014 at 11:02 history edited Ricardo Andrade CC BY-SA 3.0
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Jan 31, 2014 at 11:01 comment added Piotr Pstrągowski Oh, you're right. I believe we could first deform $i$ to be cellular, then take the mapping cyllinder, then use the homotopy extension property to make the resulting diagram commutative and not just commutative up to homotopy. But that's not really any easier than the argument in your answer.
Jan 31, 2014 at 10:49 history edited Ricardo Andrade CC BY-SA 3.0
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Jan 31, 2014 at 10:44 comment added Ricardo Andrade Dear @Piotr Pstrągowski, I am afraid I do not see why the mapping cylinder of $i$ must be a CW-complex. After all, cells of lower dimension in $X$ may attach to cells of higher dimension in $Y$. The real issue is guaranteeing $X$ is a CW-complex in the end.
Jan 31, 2014 at 10:40 comment added Piotr Pstrągowski Can't you assume that $i: Y \rightarrow X$ is an inclusion of a subcomplex by replacing $X$, if needed, by the mapping cylinder of $i$? Anyways, great answer!
Jan 31, 2014 at 10:37 history answered Ricardo Andrade CC BY-SA 3.0