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For a triple of spaces $Z\subseteq Y\subseteq X$, I'm looking for an example where $(X,Z)$ and $(X,Y)$ have the homotopy extension property, but $(Y,Z)$ doesn't.

I can come up with examples of pairs that don't have the homotopy extension property (e.g. $(0,1)\subset [0,1]$), but I can't find a way to combine it into an example of the above statement.

14159
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1 Answers1

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You won't find an example.

Proposition: Given $Z\subseteq Y\subseteq X$, if both $Y\subseteq X$ and $Z\subseteq X$ have the HEP, then $Z\subseteq Y$ has the HEP.

To prove this it will suffice to define a retraction $$r:Y\times I\rightarrow(Z\times I)\cup (Y\times 0).$$

The trick to this is as follows. As shown here, because $Y\subseteq X$ has the HEP, we can find a continuous function $u:X\rightarrow[0,1]$ such that $(i)$ $Y\subseteq u^{-1}(0)$, and $(ii)$ there is a retraction $s:U\rightarrow Y$ where $U=u^{-1}[0,1)$.

Satz 1 (Dold [1]): The inclusion $Z\subseteq U$ has the HEP. $\blacksquare$

Consequently there is a retraction $$t:U\times I\rightarrow (Z\times I)\cup(U\times 0).$$ We define $r$ to be the composite $$r:Y\times I\hookrightarrow U\times I\xrightarrow{t} (Z\times I)\cup(U\times 0)\xrightarrow{s\times 1}(Z\times I)\cup(Y\times 0).$$ This makes sense since the retraction $s$ fixes $Z\subseteq Y$ pointwise. Evidently $r$ has the required propreties.

$\blacksquare$

Remarks: I haven't assumed that either of the inclusions is closed and it is not necessary to do so. Neither the proposition nor its proof are mine.

[1] Dold, A. Die Homotopieerweiterungseigenschaft (=HEP) ist eine lokale Eigenschaft. Invent Math 6 (1968), 185–189 (1968).

Tyrone
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  • This is a cool proof, thank you for sharing! Do you know if there's a similar proof for the case when $(X,Z)$ and $(Y,Z)$ have the HEP (I'd like to know if any two pairs having HEP implies the third has it)? – 14159 Feb 02 '23 at 07:01
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    @14159 Here is a counterxample to that case: let $X=I^2$ be the unit square, let $Y=(I\times{0})\cup({0,1/n\mid n\in\mathbb{N}}\times I)$ be the comb space, and let $Z=I\times{0}$. Clearly both $(X,Z)$ and $(Y,Z)$ have the HEP. However $(X,Y)$ does not have the HEP, since the inclusion $Y\subseteq X$ is a homotopy equivalence, but not a deformation retract. – Tyrone Feb 02 '23 at 07:23