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I want to show $H_i(M, M - \{x \}) \cong H_i( \mathbb{R}^n, \mathbb{R}^n - \{0\} )$ via excision and can't quite figure out how to choose my subspaces.

For $Z \subset A \subset X$, excision gives the following isomorphism

$$ H_i(X -Z, A-Z) \cong H_i(X,A)$$

If $X=M$ is a manifold, I need $M-Z = \mathbb{R}^n$, so some ball in $M$. I was thinking of possibly letting $Z = M - B_x$ but this isn't quite working out.

Furthermore,I am completely stuck using excision to show that $H_n(M, M- \{x\}) \cong H_n(M, M - B_x)$ where $B_x$ some ball around $x$.

Any help is appreciated!

Yuugi
  • 2,143

1 Answers1

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What exactly is not working out? Let $B_x$ be the interior of a closed ball around $x$, and let $Z = M - B_x$, and $A = M - \{x\}$. Then the excision isomorphism is exactly what you're asking for: $$H_i(\Bbb R^n, \Bbb R^n \setminus \{0\}) \cong H_i(B_x, B_x - \{x\}) = H_i(M-Z,A-Z) \cong H_i(M,M - \{x\}).$$ (The first isomorphism is because the open ball is homeomorphic to $\Bbb R^n$, and I can choose that homeomorphism to take $\{x\}$ to $\{0\}$.)

As for your second question: $M - \{x\}$ deformation retracts onto $M - B_x$. (Prove this.) So indeed the pairs $(M,M-\{x\})$ and $(M,M-B_x)$ are homotopy equivalent. (As a hint for how to prove this: First prove it for $(\Bbb R^n,\Bbb R^n \setminus \{0\})$ and $(\Bbb R^n, \Bbb R^n \setminus D^n)$, $D^n$ the open disc of radius $1$.)