# Homology of Klein bottle with Mayer-Vietoris

I'm practicing with using the Mayer-Vietoris sequence, and found this computation. I thought it would be a good exercise to try cutting the Klein bottle into two cylinders, instead of into two mobius strips; i.e. as in this picture.

We then have $$A$$ and $$B$$ homotopic to $$S^1$$ and $$A\cap B$$ homotopic to $$S^1\oplus S^1$$. So we get the Mayer-Vietoris sequence with the only nonzero part being

$$0\to H_2(K)\to H_1(A\cap B)\to H_1(A)\oplus H_1(B)\to H_1(K)\to \overset{\sim}{H_0}(A\cap B)\to 0$$

which is just $$0\to H_2(K)\xrightarrow{\alpha} \mathbb{Z}\oplus \mathbb{Z}\xrightarrow{\beta} \mathbb{Z}\oplus \mathbb{Z}\xrightarrow{\gamma} H_1(K)\xrightarrow{\delta} \mathbb{Z}\to 0$$ Here $$\beta$$ sends a pair of rotations around $$S^1$$ to their inclusion in the cylinders $$A$$ and $$B$$. Since $$B$$ reverses the rotation of one of the pair (in comparison to $$A$$), we see $$\beta(x,y)=(x+y,x-y)$$. Hence $$\ker\beta=\mathrm{im}\ \alpha$$ is trivial, and so $$\alpha$$ is the zero map, thus $$H_2(K)\cong 0$$.

Next a pair $$(x,y)\in \mathrm{im}\ \beta$$ if and only if $$x-y$$ is even, which makes $$\mathrm{coker}\ \beta=\mathbb{Z}_2$$, so $$\gamma$$ induces an injective map $$\mathbb{Z}_2\xrightarrow{\gamma^*} H_1(K)$$, giving the short exact sequence $$0\to \mathbb{Z}_2\xrightarrow{\gamma^*} H_1(K)\xrightarrow{\delta} \mathbb{Z}\to0$$

The thing I'm struggling with is showing that this sequence is split, which implies $$H_1(K)\cong \mathbb Z\oplus\mathbb Z_2$$, as it should.

Am I doing this correctly? How should I continue?

• $\mathbb{Z}$ is free, hence every exact sequence $0\to A\to B \to \mathbb{Z}$ splits Commented Feb 12, 2019 at 20:04

As noted in a comment, your sequence $$0\to \mathbb{Z}/2\xrightarrow{\gamma^*} H_1(K)\xrightarrow{\delta}\mathbb{Z}\to 0$$ splits since $$\mathbb{Z}$$ is free. A splitting map for $$\delta$$ would be just $$\mathbb{Z}\to H_1(K)$$ sending $$1$$ to any preimage of $$1$$ (which exists since $$\delta$$ is surjective).