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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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    What is $V$? If it is the whole space, then trivially $V = \textrm{int}(V)$.2011-08-08
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    No, $V$ might not be $X$.2011-08-08
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    What about a point $\{x\} = A$ lying on the boundary of a closed ball $V$?2011-08-08
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    When you write ball you mean disc, i.e. $D^n$, not $S^n$, right? Then the closure of the point is not in the interior so that is a counter example. Now I'm confused because on p 124 Hatcher applies the excision theorem but $\bar{A} \subset int(V)$ doesn't necessarily hold : (2011-08-08
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    Are you talking about Proposition 2.22? There $V$ is a *neighborhood* of $A$ in $X$ and $A$ is closed. Yes, I meant a closed disk, not a sphere.2011-08-08
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    Oh. It has to be closed too, not just a deformation retract! You're right. Thank you!2011-08-08
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    @Matt: It is standard to use ‘ball’ for the solid object and ‘sphere’ for the hollow object. [Redundant comments deleted.]2011-08-08
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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 $\bar{A} \subset int(V)$.