9
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ I am doing my first course in algebraic (or in fact any) topology, and I often come across your questions on this site. By now it seems to me that you are like an expert in this field... Having done other topics in maths, I find this area unusually painful, and so I was just wondering how you found this journey? Any tips on resources besides Hatcher? $\endgroup$
    – gen
    Oct 29, 2018 at 21:13
  • 1
    $\begingroup$ Why is $A\cap B\simeq \mathbb{S}^1$? According to the wikipedia picture we should have $A\cap B\simeq \mathbb{S}^1\sqcup\mathbb{S}^1$, shouldn't we? $\endgroup$
    – rmdmc89
    Oct 6, 2019 at 16:23

1 Answer 1

8
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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 $2 \mathbb{Z} (1,1) = \mathbb{Z} \oplus 2\mathbb{Z}$... $\endgroup$ Aug 21, 2011 at 17:54
  • 4
    $\begingroup$ 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 $n \in 2\mathbb Z$. $\endgroup$ Aug 21, 2011 at 18:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .