4
$\begingroup$

Let $X$ be a topological space and $x \in X$ be such that $\{x\}$ is closed in $X$. Let $y \in \mathbb{R}^n$. I've read that $$ H_k(\mathbb{R}^n \times X, (\mathbb{R}^n \times X) - \{(y,x)\}) \cong H_{k - n}(X,X - \{x\}). $$ for every $k$. I have no clue how to prove that. Can someone help?

$\endgroup$
0
$\begingroup$

First let me say that I'm really no expert on algebraic topology, only because there still wasn't an answer up to now I thought it might be helpful to give my suggestion for a proof anyways: Let us first assume $X$ is a CW complex. Further I assume you are referring to real coefficients (though this doesn't matter too much for the following).

  • From e.g. Hatchers book "Algebraic Topology", section 3.B, we know that there is an exact sequence of relative homologies, where we set $A:=X\setminus\{x\}$ and $B:=\mathbb{R}^n\setminus \{y\}$: $$\tag{1}0\rightarrow\oplus_i(H_i(X,A)\otimes_{\mathbb{R}}H_{k-i}(\mathbb{R}^n,B))\overset{(*)}{\rightarrow }H_k(X\times\mathbb{R}^n,A\times \mathbb{R}^n\cup X\times B)\rightarrow\oplus_i\mathrm{Tor}_{\mathbb{R}}(H_i(X,A),H_{k-i-1}(\mathbb{R}^n,B))\rightarrow 0 $$

  • Now, we use the long exact homology sequence (cf. section 2.1 of Hatchers book) $$\ldots\rightarrow H_i(B)\rightarrow H_i(\mathbb{R}^n)\rightarrow H_i(\mathbb{R}^n,B)\rightarrow H_{i-1}(B)\rightarrow\ldots$$ to prove that $H_i(\mathbb{R}^n,B)$ is $\mathbb{R}$ if $i=n$ and $0$ else (where we use the fact that the homology groups are invariant under homotopy and $\mathbb{R}^n$ is homotopic to a point, $B=\mathbb{R}^n\setminus\{y\}$ is homotopic to a sphere $S^{n-1}$). In particular, the Tor term in the exact sequence (1) vanisches, i.e. the morphism ($*$) is an isomorphism, and the direct sum of the first factor collapses to $H_{k-n}(X,A)\otimes_{\mathbb{R}}\mathbb{R}\simeq H_{k-n}(X,A)$.

  • Observing that $A\times\mathbb{R}^n\cup X\times B=X\times \mathbb{R}^n\setminus(\{x\}\times\{y\})$ finishes the prove.

To get rid of the assumption that $X$ is a CW complex you might want to read the paragraph in Hatchers book about CW approximation (in section 4.1), but I really don't know about this, I just have the vague idea it might help (sorry if it doesn't)...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy