I would guess, I have to use excision theorem. But I don't see, how i see $\mathbb{D}^2$ as a subset of $F$. Neither, there is a homeomorphism between $\mathbb{D}^2$ and $F$ for an arbitrary surface. Any suggestions?
How do I show $H_n (F, F - {x} ) \cong H_n ( \mathbb{D}^2 , \mathbb{D}^2 - 0) $? for a surface F.
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algebraic-topology
homology-cohomology
surfaces
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2Well, what is the definition of a surface? – 2017-01-16
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0A hausdorff space, that is locally homeomorphic to $\mathbb{R}^2$. One can show that for any neighbourhood $U$ of $x$ $H_n (F, F - x) \cong H_n (U, U- x )$ So, one gets $H_n (F, F - x) \cong H_n (\mathbb{R}^2 , \mathbb{R}^2 - 0 )$. With excision theorem, one gets $H_n (\mathbb{R}^2 , \mathbb{R}^2 - 0 ) \cong H_n (\mathbb{D}^2 , \mathbb{D}^2 - 0 ) $ Thanks for your suggestion to look up what a surface is again.^^ – 2017-01-16
1 Answers
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One can show that for any neighbourhood $U$ of $x$ $H_n (F, F - x) \cong H_n (U, U- x )$ So, one gets $H_n (F, F - x) \cong H_n (\mathbb{R}^2 , \mathbb{R}^2 - 0 )$. With excision theorem, one gets $H_n (\mathbb{R}^2 , \mathbb{R}^2 - 0 ) \cong H_n (\mathbb{D}^2 , \mathbb{D}^2 - 0 ) $