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Let $X$ be a finite dimensional CW complex and $A$ be a closed subset in $X$ and $N$ a regular neighborhood of $A$ that deformation retracts onto it. why do we have for each $i$,

$$H^{i}(X-A;\mathbb Z)\cong H^{i}(X-N;\mathbb Z)$$

My guess: if two subspaces $A$ and $B$ of $X$ are homotopy equivalent then their complements $X-A$ and $X-B$ must be homotopy equivalent and then have the same cohomology groups?

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    Do you have some kind of theorem like excision for cohomology? Maybe you can use that.2012-10-01
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    Excision is about relative cohomology which is not the case here.2012-10-01
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    If am sure you can relate the relative cohomology of say $H^n(X,\emptyset;\Bbb{Z})$ to the cohomology of just $X$...2012-10-01
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    You **do** have a pair, though: $(X-A,X-N)$. Use the long exact sequence for that!2012-10-02
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    Your guess, by the way, is wrong: two spaces can be in fact homeomorphic without their complements in a bigger space being homotopy equivalent (nor even having the same homology). Google for «Alexander's horned sphere». (If you write what you really meant to write, then... :-) )2012-10-02
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    @MarianoSuárez-Alvarez : Writing the long exact sequence of the pair, I see that if for each $i$, $H^i(X-A,X-N)=0$ then $H^i(X-A)\cong H^i(X-N)$ but why could $H^i(X-A,X-N)=0$ be true for each $i$?2012-10-02
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    You have to use the fact that you have a deformation retract at some point... Think about this for a while. In a couple of days I'll add an answer.2012-10-02
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    Is the following true: if $N$ deformation retracts onto $A$ then $X-A$ deformation retracts onto $X-N$?2012-10-11

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