I'm still applying Mayer Vietoris, this time to the Klein bottle. I'm using the decomposition as on Wikipedia here and I've calculated $H_0$ and $H_n$ for $n \geq 2$ correctly. Now I'm struggling with $H_1$.
My sequence:
$$ 0 \xrightarrow{} H_1( S^1) \xrightarrow{(i,j)} H_1(M) \oplus H_1(M^\prime )\xrightarrow{k-l} H_1(K) \xrightarrow{\partial_1} \tilde{H_0} = 0 $$
Wikipedia writes "The central map $(i,j)$ sends $1$ to $(2, −2)$". Does it matter whether it's sent to $(2,-2)$ or $(2,2)$ ? I think $(i,j)$ maps $1$ to $(2,2)$.
Using this, I get
(i) $im((i,j)) = 2 \mathbb{Z} \oplus 2 \mathbb{Z} = ker (k-l)$
And using the first isomorphism theorem I get
(ii) $H_1(K) / ker (\partial_1) = im(\partial_1) = \tilde{H_0} = 0$
and therefore $H_1(K) \cong \mathbb{Z}$ because $ker(\partial_1) = \mathbb{Z}$ which is clearly wrong but I don't see where the mistake is.
(iii) I also know $k-l$ is surjective so $im (k-l) = H_1(K) \cong \mathbb{Z} \oplus \mathbb{Z} / ker (k-l)$
But if $ker (k-l) = 2 \mathbb{Z} \oplus 2 \mathbb{Z} $ then I'd get $H_1(K) = \mathbb{Z}/2 \oplus \mathbb{Z}/2 $
which is also wrong.
What am I doing wrong? Many thanks for your help!
Best Answer
Where $(i,j)$ sends $1$ depends on how your inclusion maps look and which orientation you pick, that is which isomorphisms $H_1(S^1) \cong \mathbb Z$, $H_1(M) \cong \mathbb Z$ and $H_1(M') \cong \mathbb Z$ you pick. It is therefore ok to assume $(i,j)1 = (2,2)$.
Now you get
(1) $im(i,j) = \ker (k-l) = 2\mathbb Z(1,1) \not = 2\mathbb Z \times 2 \mathbb Z$.
(2) $\partial_1 = 0$ and therefore $k-l$ is surjective. We now have $H_1(K) \cong [\mathbb Z \times \mathbb Z] / [2\mathbb Z(1,1)]$.
(3) Use that $\mathbb Z \times \mathbb Z = \mathbb Z (1,0) \oplus \mathbb Z(1,1)$ to conclude $H_1(K) \cong \mathbb Z \times \mathbb Z/2\mathbb Z$.