Relative homology $H_n(T^2,S^1)$ (Hatcher, Exc 2.1,17.b) 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).

 A: Yes. Since $\mathbb Z$ is free, the sequence splits and you get what you want. 
A: In case you're not aware of the splitting lemma, let's prove this by hand (this proof obviously extends easily to the fully general case).
First, let $H = \tilde{H}_1(T,S)$ and let's give the maps names:
$$0\rightarrow \mathbb{Z}\oplus\mathbb{Z} \overset{f_1}{\rightarrow} H \overset{f_2}{\rightarrow}\mathbb{Z}\rightarrow 0.$$
Note that the map $\mathbb Z \to 0$ is trivial so its kernel is the whole of $\mathbb{Z}$. Hence, the map $H \overset{f_2}{\to} \mathbb Z$ must be surjective by exactness. Let $h \in H$ be an element such that $f_2(h) = 1$. Define the function $g \colon \mathbb{Z} \to H$ on the generator by $g(1) = h$. It is then easy to see that $f_2(g(n)) = n$ for all $n \in \mathbb{Z}$.
Now define a map $i \colon (\mathbb{Z}\oplus \mathbb{Z}) \oplus \mathbb{Z}  \to H$ by $i(x,y) = f_1(x) + g(y)$. We see that $i$ is clearly a homormophism.
Note that if $i(x,y) = 0$ then $f_1(x) + g(y) = 0$ which means that $f_1(x)=g(y)$ and so $f_2(f_1(x)) = f_2(g(y))$. But $f_2(g(y))=y$ by the definition of $g$. Also, by exactness, $f_2(f_1(x)) = 0$, so we have $y=0$. It follows that $f_1(x) = 0$ and, as $f_1$ is injective by exactness, $x$ must be $0$. It follows that $i$ is injective.
Now, let $z \in H$. Let $\tilde{z} = z - g(f_2(z))$. We have $f_2(\tilde{z}) = f_2(z) - f_2(g(f_2(z))) = f_2(z) - f_2(z) = 0$ so $\tilde{z} \in \ker f_2$. It follows that $\tilde{z} \in \operatorname{im}f_1$ by exactness. Let $a \in \mathbb{Z} \oplus \mathbb{Z}$ be such that $f_1(a) = \tilde{z}$. Finally, we see that
$$i(a,f_2(z)) = f_1(a)+g(f_2(z)) = \tilde{z} + g(f_2(z)) = z - g(f_2(z)) +g(f_2(z)) = z.$$ Hence, $z$ is in the image of $i$ and so $i$ is surjective. It follows that $i$ is an isomorphism.
