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If $A$ is a deformation retract of $V$, does it follow that $\bar{A} \subset \operatorname{int}(V)$? If yes, how? I think Hatcher uses it implicitly.

Many thanks for your help.

Edit: The spaces look like this: $A \subset V \subset X$

Edit 2: I'm trying to apply excision to a good pair.

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    Yes, Prop. 2.22 it is. But now it's clear, I had written down the definition of good and omitted that it had to be closed. If $V$ is a neighbourhood of$A$and A is closed then $\bar{A} \subset A \subset int(V)$.2011-08-08

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A good pair is a pair $(X,A)$ s.t. $A$ closed in $X$ and $\exists$ neighbourhood $V$ of $A$ s.t. is a deformation retract of $V$.

Then clearly $A = \bar{A} \subset O \subset int(V) \subset V$ where $O$ open in $X$ and therefore $\bar{A} \subset int(V)$.