Using the five lemma in proving $H_{n}(D^k, D^k \setminus \{0\} ) \cong H_{n}(D^k, S^{k - 1}).$

Question

I want to show that for all $n,$ $H_{n}(D^k, D^k \setminus \{0\} ) \cong H_{n}(D^k, S^{k-1} ).$

My trial is:

1-I know that $D^k \setminus \{0\}$ is homotopically equivalent to S^{k-1}, but I do not know how this is helpful here. Also, I do not know how to create the two rows of exact sequences of abelian groups and why my diagram will commute if I am going to use the 5-lemma, could anyone explain this for me, please?

Could anyone help me answer those questions?

2- Also it is not clear for me when should I think that I should create a long exact sequence and when should I think that I should apply the 5-lemma?

Answer 1

The only tools that we have in basic topology for dealing with relative homology groups are the long exact sequence of a pair and for good pairs the isomorphism $H_n(X,A)\cong \bar{H}(X/A) $. You are interested in the relationship between two relative homology groups and there is an obvious relative map between the two. Relative maps give maps of the long exact sequence of a pair. This would be a good thing to apply the five lemma to.

Answer 2

You probably know that we have a long exact sequence of the form:

$... \rightarrow H_n(A) \rightarrow H_n(X) \rightarrow H_n(X,A) \rightarrow H_{n-1}(A) \rightarrow H_{n-1}(X)\rightarrow ...$

You probably know that $D^{k} - \{0\}$ is homotopy equivalent to $S^{k-1}$, and therefore $H_n(D^{k} - \{0\}) \cong H_n(S^{k-1})$.

So you have two long exact sequences:

$... \rightarrow H_n(D^{k}-\{0\}) \rightarrow H_n(D^{k}) \rightarrow H_n(D^{k},D^{k}-\{0\}) \rightarrow H_{n-1}(D^{k} - \{0\}) \rightarrow H_{n-1}(D^{k})\rightarrow ...$

$... \rightarrow H_n(S^{k-1}) \rightarrow H_n(D^{k}) \rightarrow H_n(D^{k},S^{k-1}) \rightarrow H_{n-1}(S^{k-1}) \rightarrow H_{n-1}(D^{k})\rightarrow ...$

The outwards arrows are isomorphisms, therefore by the $5$ lemma, so is $H_n(D^{k},S^{k-1}) \cong H_n(D^{k},D^{k} - \{0\}) $