Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 4 down vote accepted

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 $H_1(K) \cong \mathbb Z \times \mathbb Z/2\mathbb Z$.

share|cite|improve this answer
Thank you! But what is $2 \mathbb{Z} (1,1)$? First I thought it's the free abelian group over the basis consisting of one element, $(1,1)$ but then that would be $\mathbb{Z}$ and I need it to be $\mathbb{Z} \oplus 2\mathbb{Z}$ but I don't see how $2 \mathbb{Z} (1,1) = \mathbb{Z} \oplus 2\mathbb{Z}$... – Rudy the Reindeer Aug 21 '11 at 17:54
You should read this with some linear algebra in mind. $2\mathbb Z (1,1)$ is the subgroup of $\mathbb Z \times \mathbb Z$ that consists of the elements $(n,n)$ for some $n \in 2\mathbb Z$. – Alexander Thumm Aug 21 '11 at 18:02
Thank you! .... – Rudy the Reindeer Aug 22 '11 at 6:52
You're welcome! – Alexander Thumm Aug 22 '11 at 11:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.