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Let $X$ be a compact, oriented, connected $n$-manifold without boundary. Let $D$ be a $n$-disk in some chart domain of $X$. Then $X-D\simeq X-\{p\}$ is a non-compact, oriented, connected manifold without boundary, and $X-$int$(D)$ is a compact, oriented, connected manifold with boundary $\partial X\simeq S^{n-1}$.

I read in many places that $X-D\simeq X-\{p\}$ has top cohomology group zero (here). As for $X-$int$(D)$, Bredon's Topology and Geometry's page 356 hasenter image description here where $M$ is a compact, connected, oriented manifold with boundary $\partial X$. This means $\mathrm{H}^n(X-\text{int}(D))\simeq \mathrm{H}_0(X-\text{int}(D),S^{n-1})$, which is nonzero.

But since int$(D)$ can retract to a smaller closed disk $D'$, we should have $X-D$ is homotopic to $X-$int$(D)$, and their cohomology groups isomorphic. Where did I go wrong?

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  • $\begingroup$ Why should $X\setminus D\simeq X\setminus int(D)$? I don't think this is true in general. $\endgroup$ – Vincent Boelens Dec 14 '16 at 16:02
  • $\begingroup$ @VincentBoelens if $X$ is a manifold, this is true, no? assume $D$ is contained in a chart domain. then that result follows from the fact that $S^{n-1}\times [0,1)\simeq S^{n-1}$. $\endgroup$ – zudumathics Dec 14 '16 at 17:32
  • $\begingroup$ by $\simeq$ I mean homotopic $\endgroup$ – zudumathics Dec 14 '16 at 17:32
  • $\begingroup$ $X-int(D)$ is connected. The copy of $S^{n-1}$ is in that connected component. What do you think that makes $\mathrm{H}_0(X-\text{int}(D),S^{n-1})$? If you like, you can use the long exact sequence of a pair. $\endgroup$ – PVAL-inactive Dec 14 '16 at 17:57
  • $\begingroup$ @zudumathics I am not convinced. For chart domains this works, but how do you extend it to $X$? $\endgroup$ – Vincent Boelens Dec 15 '16 at 10:13

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