Exact sequence and torsion I've come across another exact sequence, where (I guess) I need to deduce the result using some properties of torsion.
I am calculating the homology of the Klein bottle using attaching maps.
I start by defining $\Phi:I \times I \to K$ as the natural map and denote $\partial(I \times I)$ as the boundary, then let $f=\Phi|\partial(I \times I)$. We can then regard $f$ as a function $S^1 \to S^1 \times S^1$. 
I think I can show that the induced map $f_*: H_1(S^1) \to H_1(S^1 \vee S^1)$ has degree 2 (i.e. is multiplication by 2)
It boils down to the following exact sequence and have come across the following exact sequence (I am trying to calculate $H_1(K)$)
$$0 \to \mathbb{Z} \stackrel{f_*}{\to} \mathbb{Z} \oplus \mathbb{Z} \stackrel{i_*}{\to} H_1(K) \to \mathbb{Z}$$
I know that $H_1(K)$ must have rank 1 (from the Euler characteristic of the Klein bottle)
I note that previously when I had a sequence $H_1(S^1 \vee S^1) \to H_1(T) \to H_0(S^1)$ the book concluded that $H_1(T)$ was torsion free (here $T$ is the torus), but as Jim pointed out to me, $H_1(K)$ is not torsion free this time. 

The composition of the map $f:S^1\to S^1\vee S^1$ with one of the projections $S^1\vee S^1\to S^1$ is a map $S^1\to S^1$ of degree two: indeed, this composition turns around the codomain $S^1$ twice. Now it is easy to see that if $p_1$, $p_2:S^1\vee S^1\to S^1$ are the projections, then the map $$\left(\begin{array}{c}p_{1*}\\p_{2*}\end{array}\right):H_1(S^1\vee S^1)\to H_1(S^1)\oplus H_1(S^1)$$ is an isomorphism. It follows that the composition $$\left(\begin{array}{c}p_{1*}\\p_{2*}\end{array}\right)\circ f_*$$ is, in matrix terms, $$\left(\begin{array}{c}2\\2\end{array}\right).$$
It follows that your exact sequence is 
$$0 \to \mathbb{Z} \stackrel{\left(\begin{array}{c}2\\2\end{array}\right)}{\to} \mathbb{Z} \oplus \mathbb{Z} \stackrel{i_*}{\to} H_1(K) \to \mathbb{Z}$$
