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Let $X$ be an arbitrary topological space and $A$ a subspace such that there exists a (strong) deformation retraction of $X$ to $A$.

Does it follow that $(X,A)$ has the homotopy extension property? If not, what are some nice counter-examples?

I suspect this might be a helpful fact to know when solving various exercises in algebraic topology. I can't think of an easy argument in any direction, so any thoughts would be welcome.

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up vote 3 down vote accepted

This is almost true. What you need to additionally assume is that $A$ is a zero set, i.e. $A=\varphi^{-1}[\{0\}]$ for some continuous $\varphi\colon X\to [0,1]$. Since your $A$ is presumably closed, this will always be true in reasonable spaces (a strong separation axiom is needed which is true e.g. in metric spaces or locally finite CW complexes).

There is the following characterization: if $A\subseteq X$ is closed, then $(X,A)$ has the homotopy extension property iff $A$ is a zero set and some neighbourhood of $A$ strongly deformation retracts onto it. This is theorem 1 of this. It is also covered in May's A Concise Course

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Thanks. Do you know of any interesting counter-examples perhaps? (Non-reasonable spaces and non-closed $A$ are fine.) –  Dejan Govc Nov 17 '12 at 22:18
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@DejanGovc I guess a counterexample would be a deformation retract of a space which isn't a zero set? Note that zero sets are $G_\delta$, so some sort of failure of first countability would do. In particular, take $X=[0,1]^{\mathbb{R}}$ with the product topology and $A=\{0\}$ the singleton of the constant function. There's an obvious deformation retraction but $A$ isn't a zero set so isn't cofibrantly included (note that it is still closed though). –  Miha Habič Nov 17 '12 at 22:35
    
Beautiful, thanks! –  Dejan Govc Nov 17 '12 at 22:40
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