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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 $V = \textrm{int}(V)$. – Zhen Lin Aug 8 '11 at 8:39
No, $V$ might not be $X$. – Rudy the Reindeer Aug 8 '11 at 8:42
What about a point $\{x\} = A$ lying on the boundary of a closed ball $V$? – t.b. Aug 8 '11 at 8:43
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 : ( – Rudy the Reindeer Aug 8 '11 at 8:48
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. – t.b. Aug 8 '11 at 8:51
up vote 2 down vote accepted

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 $\bar{A} \subset int(V)$.

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