I am going through Hatcher's Algebraic Topology and I am trying to solve exercise 17.b of part 2.1. The question asks to compute the relative homology $H_n(X,B)$.
As far as I can understand , if on a double torus we collapse the $B$ circle, it is the same as collapsing a set $S=\{pt_1,pt_2\}$ of two points on a a regular torus $T$.
For that I am going to use the long exact sequence of reduced homology (Theorem 2.13 in Hatcher). By that theorem we get:
$\tilde{H}_1(S)\rightarrow \tilde{H_1}(T)\rightarrow \tilde{H}_1(T,S)\rightarrow \tilde{H}_0(S)$.
Now, this gives us the exact sequence $0\rightarrow \mathbb{Z}\oplus\mathbb{Z}\rightarrow \tilde{H}_1(T,S)\rightarrow\mathbb{Z}\rightarrow 0$.
Is there any "easy" way to deduce that the homology group is $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}$? (this is what my intuition tells me).
Best Answer
Yes. Since $\mathbb Z$ is free, the sequence splits and you get what you want.