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.
On
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 $1$.
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?