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Let $X$ be a closed $n$-manifold, $B$ an open $n$-disc in $M$. Suppose $p:X\rightarrow X/(X-B)$ is a quotient map. Notice that $X/(X-B)$ is homeomorphic to the sphere $\mathbb{S}^n$.

My question is whether the quotient map induces an isomorphism mapping $H^n(X;\mathbb{Z}_2)$ to $H^n(X/(X-B);\mathbb{Z}_2)$? Maybe it is known. But I can not find it in the literature. If it is easy, can somebody show it to me? Thanks a lot.

Lewis Zhang
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2 Answers2

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Consider the map of pairs $(X,X-B)\to(S^n,*)$ with $S^n=X/(X-B)$ and $*$ the point resulting from collapsing $X-B$, and the diagram you get from it in the long exact sequences for cohomology. The map $H^n(S^n,*)\to H^n(S^n)$ is an isomorphism, and so is $H^n(S^n,*)\to H^n(X,X-B)$, so by the commutativity of one of the sqaures in the diagram you want to know if $H^n(X,X-B)\to H^n(X)$ is an isomorphism. Is it?

Mariano Suárez-Álvarez
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  • In fact, I don't know why $H^n(S^n,∗)\rightarrow H^n(X,X−B)$ and $H^n(X,X−B)\rightarrow H^n(X)$ are isomorphisms. Can you show it to me? – Lewis Zhang Jan 29 '15 at 03:28
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    The first one you can get using excision and a little work. The second one, by using orientations and their properties. – Mariano Suárez-Álvarez Jan 29 '15 at 04:52
  • Thanks. I think the first one also depends on the fact that $X-B$ is a deformation retract of a open subspace of $X$. Since I use the coefficient $\mathbb{Z}_2$, the second one follow from $H^n(X\setminus B)=0$, which I don't know why. – Lewis Zhang Jan 29 '15 at 13:44
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    I did not say that the 1st one depends only on excision: what you mention is the little work I referred to, but the basic fact is that you can use excision to see that the relative groups involved do not depend much on what's inside of the X-B. As I daid, the 2nd one you can get by shoing that you can extend orientations to the whole of B; see the treatment of orientability on manifolds in the book by Greenberg and Harper. – Mariano Suárez-Álvarez Jan 29 '15 at 19:27
  • OK, thank you for your comment! – Lewis Zhang Jan 30 '15 at 01:30
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Note that $X\setminus B$ deformation retracts onto a lower dimensional subspace (assuming $X$ is connected) and so in particular $H^n(X\setminus B)=0$. Consider the long exact sequence in cohomology of the pair $(X,X\setminus B)$ given by $$\ldots\to H^n(X/(X\setminus B))\to H^n(X)\to H^n(X\setminus B)\to 0\to \ldots$$

As $H^n(X\setminus B)=0$ we can conclude that $q^*\colon H^n(X/(X\setminus B))\to H^n(X)$ is surjective and so also an isomorphism as $H^n(X)$ and $H^n(S^n)$ both have rank $1$.

Dan Rust
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  • The claim that the complement of $B$ deformation retracts onto a lower dimensional subspace is somewhat nontrivial, no? – Mariano Suárez-Álvarez Jan 28 '15 at 18:02
  • @MarianoSuárez-Alvarez You're right. In fact I would probably use a homology argument to prove that. I guess that makes the above argument a little circular... – Dan Rust Jan 28 '15 at 18:07
  • Where is the fact "$X\setminus B$ deformation retracts onto a lower dimensional subspace" come from? – Lewis Zhang Jan 29 '15 at 13:09
  • @LewisZhang, as we observed in the comments above yours, that fact is in fact probably more difficult than the one you want to prove. – Mariano Suárez-Álvarez Jan 30 '15 at 03:37
  • @LewisZhang as Mariano points out, proving $H^n(X\setminus B)=0$ is at least as difficult, either by having to prove $X\setminus B$ deformation retracts to a lower dimension, or by considering a relative LES in cohomology for the pair $(X\setminus B,\partial B)$ which would then just push the problem back another step. In the orientable case you can use for instance the answer to this question, but I would suggest awarding the accepted answer to a more complete one than mine. – Dan Rust Jan 30 '15 at 11:02
  • I'll try to fix this argument for the general case, but if I can't I'll delete this answer. – Dan Rust Jan 30 '15 at 11:03