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?

  • 2
    $\begingroup$ $\mathbb{Z}$ is free, hence every exact sequence $0\to A\to B \to \mathbb{Z}$ splits $\endgroup$ Feb 12, 2019 at 20:04

1 Answer 1


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).


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