I am kind of lacking of intuition for questions like that. I see that a chain in $\mathbb{D}^2 - 0$ can be continuously transformed into a chain in $\partial \mathbb{D}^2$, but with the relative homology, I have problems to understand.
How do I show $ H_n (\mathbb{D}^2 , \mathbb{D}^2 - 0 ) \cong H_n ( \mathbb{D}^2 , \partial \mathbb{D}^2 )$?
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algebraic-topology
homology-cohomology
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0Observe that $\mathbb{D}^2-0$ is homotopy equivalent to the boundary of disk. – 2017-01-17
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0I saw that, but the relative homology is defined as the Homology of $S_n ( \mathbb{D}^2 / S_n ( \mathbb{D}^2 - 0 )$ and so I cannot apply directly homotopy invariance. – 2017-01-17
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0What tools do you have available? Could you use long exact sequences, homotopy invariance, or the 5 lemma? – 2017-01-17
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0All of the things you mentioned, i guess. – 2017-01-17
1 Answers
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We know that $D^n-0$ is homotopic to $\partial D^n$, but you're right, this doesn't follow immediately from excision.
Take the long exact sequences for the pairs $(D^n, \partial D^n)$ and $(D^n, D^n-0)$. Since we have an inclusion $\partial D^n\hookrightarrow D^n-0$ which is a restriction of the identity on $D^n$, then we have a map from our $(D^n, \partial D^n)$ long exact sequence of homology groups to our $(D^n, D^n-0)$ sequence of homology groups.
$$\begin{array} A H_k(D^n) & \stackrel{}{\longrightarrow} & H_k(\partial D^n) & \stackrel{}{\longrightarrow} & H_k(D^n,\partial D^n) & \stackrel{}{\longrightarrow} & H_{k-1}(D^n) & \stackrel{}{\longrightarrow} & H_{k-1}(\partial D^n)\\ \downarrow{id_\ast} & & \downarrow{i_\ast} && \downarrow{f} & & \downarrow{id_\ast} & & \downarrow{i_\ast} \\ H_k(D^n) & \stackrel{}{\longrightarrow} & H_k(D^n-0) & \stackrel{}{\longrightarrow} & H_k(D^n,D^n-0) & \stackrel{}{\longrightarrow} & H_{k-1}(D^n) & \stackrel{}{\longrightarrow} & H_{k-1}(D^n-0)\\ \end{array}$$
The reason we have this map between exact sequences is purely formal, but we need to use the fact that our inclusion is a restriction of the identity to make sure all our little squares commute. Notice we obtain an induced map $f:H_k(D^n, \partial D^n)\rightarrow H_k(D^n, D^n-0)$ for all $k$. The two maps to the left and the right of $f$ are isomorphisms on homology, because the identity induces an isomorphism (obviously), and the inclusion $i:\partial D^n\hookrightarrow D^n-0$ is a homotopy equivalence. We then use the five-lemma to conclude that $f$ is an isomorphism.
Let $n=2$ and we're done. Notice that this argument only needed $D^n-0$ to deformation retract onto $\partial D^n$, so there is an obvious generalisation there.