# cohomology isomorphism

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. –  user38268 Oct 1 '12 at 15:45
Excision is about relative cohomology which is not the case here. –  palio Oct 1 '12 at 15:48
If am sure you can relate the relative cohomology of say $H^n(X,\emptyset;\Bbb{Z})$ to the cohomology of just $X$... –  user38268 Oct 1 '12 at 15:51
You do have a pair, though: $(X-A,X-N)$. Use the long exact sequence for that! –  Mariano Suárez-Alvarez Oct 2 '12 at 6:07
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... :-) ) –  Mariano Suárez-Alvarez Oct 2 '12 at 6:08