# 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.

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Your statements are not consistent. How can it be torsion-free and also contain $\mathbb Z/2\mathbb Z$? Also, $\mathbb Z\oplus\mathbb Z/2\mathbb Z\oplus \mathbb Z/2\mathbb Z$ is of rank 1 too. –  Grumpy Parsnip Mar 25 '11 at 2:03
Also, the given exact sequence is not enough to determine $H_1$. At the very least, $f_*$ needs to be determined. –  Grumpy Parsnip Mar 25 '11 at 2:04
@Jim - sorry, some mistakes. I will try and edit the question –  Juan S Mar 25 '11 at 2:07
it does not make sense to say that $f_*$ is multiplication by $2$, as it is not an endomorphism... –  Mariano Suárez-Alvarez Mar 25 '11 at 3:33
@Mariano - right again! (I am not having a great day). I think what I meant was compose $f$ with either projection $S^1 \vee S^1 \to S^1$ to give a map $\phi: S^1 \to S^1$ (which I am not so sure has degree 2 anymore - I think that might be the projective plane). I was going to use a degree argument to try and calculate the induced map $f_*:H_1(S^1) \to H_1(S^1 \vee S^!)$ - which is really the crux of the problem –  Juan S Mar 25 '11 at 3:45

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 $$0 \to \mathbb{Z} \stackrel{\left(\begin{array}{c}2\\2\end{array}\right)}{\to} \mathbb{Z} \oplus \mathbb{Z} \stackrel{i_*}{\to} H_1(K) \to \mathbb{Z}$$
@Mariano - thank you. So how can we use that to determine $H_1(K)$? I can at least see that the map $i_*$ has kernel $2\mathbb{Z} \oplus 2\mathbb{Z}$ and that it must have rank 1. I am still confused by the earlier statement (in the torus example) that $H_1(T)$ must be torsion free, and why that seemingly does not apply here –  Juan S Mar 25 '11 at 4:10
@Qwirk: the TeX was not rendering correctly, but it seems to work now... In the case of the torus, the map induced by the corresponding $f$ is rather different, so the end result is itself different! By the way: the kernel of $i_*$ is not $2\mathbb Z\oplus 2\mathbb Z$, but the submodule of $\mathbb Z\oplus \mathbb Z$ generated by $(2,2)$. That's where the torsion comes from: this means that $H_1$ has a copy of $\mathbb Z\oplus \mathbb Z/((2,2))$ which has torsion, as the class of $(1,1)$ in this quotient shows. –  Mariano Suárez-Alvarez Mar 25 '11 at 4:36