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I was wondering whether the following was true: Assume we have a pushout square $$\require{AMScd}\begin{CD} A @>>> B \\ @VVV @VVV \\ X @>>> Y \end{CD}$$ with $Y$ being the pushout of $X \leftarrow A \rightarrow B$ and the left arrow is the inclusion of a good pair $(X,A)$, that is, $A \subseteq X$ is closed and there exists an open neighborhood $A \subseteq U \subseteq X$ such that the inclusion $i\colon A \to U$ has a retraction $r \colon U \to A$ such that $r \circ i = \mathrm{id}_A$ and $i\circ r$ is homotopic to $\mathrm{id}_U$ via a homotopy $h \colon U \times [0,1] \to U$ satisfying $h(a,t) = a$ for each $a \in A$ and $t \in [0,1]$.

Does it then follow that the right arrow of the pushout square is the inclusion as above, i.e. is $(Y,B)$ a good pair?

LSpice
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Max
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1 Answers1

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Yes.

To set notation, let $U\subseteq X$ be an open neighbourhood of $A$, $r:U\rightarrow A$ a retraction, and $h:U\times I\rightarrow U$ a homotopy $ir\simeq id_U$ rel $A$, here $i:A\subseteq U$ is the inclusion.

Now form the adjunction space $$V=U\cup_AB.$$ Then $B$ is a closed subspace of $V$ and $V$ embeds into $X\cup_AB$ as an open subspace. We obtain a retraction $R:V\rightarrow B$ as the mapping induced by the composite $$U\xrightarrow{r}A\rightarrow B$$ together with the identity map of $B$. Finally, since $I$ is locally compact, $V\times I$ is the pushout of $$B\times I\leftarrow A\times I\xrightarrow{i\times1} U\times I.$$ The maps $$U\times I\xrightarrow{h}U\rightarrow V$$ and $$B\times I\xrightarrow{pr_B}B\xrightarrow{\subseteq}V$$ agree on $A\times I$, since $h|_{A\times I}=pr_A$. Let $H:V\times I\rightarrow V$ be the map supplied by the universal property of the pushout applied to the two maps above. Then $H(z,0)=R(z)$ and $H(z,1)=z$ for all $z\in V$, and moreover $H(b,t)=b$ for all $b\in B$, $t\in I$.

In particular, $(V,R,H)$ is the data of a good neighbourhood of $B$ in $X\cup_AB$.

Tyrone
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