$0 \rightarrow \mathbb{Z} \stackrel{f}{\rightarrow} \mathbb{Z}^3 \stackrel{h}{\rightarrow} H_1(X) \rightarrow 0$

Where $f(1) = (1,0,2)$

Then $H_1(X) \cong \frac{\mathbb{Z}^3}{im(f)} \cong \frac{\mathbb{Z}^3}{\langle 1,0,2 \rangle}$

Is this all correct so far? The image of $1$ under $f$ is $(1,0,2)$, but i'm just confused. The correct answer is supposed to be $H_1(X) \cong \mathbb{Z} \oplus \mathbb{Z}_2$.. I'm trying to think about the best way to make sense of this. I often have problems on steps like this when analyzing a short exact sequence. Maybe I should think of $im(g) = \mathbb{Z} \oplus 2\mathbb{Z}$ and then $\frac{\mathbb{Z}^3}{\mathbb{Z} \oplus 2\mathbb{Z}} \cong \frac{\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}}{\mathbb{Z} \oplus 2\mathbb{Z}} \cong \mathbb{Z} \oplus \mathbb{Z}_2$

Any insight in general is appreciated!! Thanks!

  • $\begingroup$ That can't be right. The image is not $\Bbb{Z}\oplus 2\Bbb{Z}$. Also according to the ses you've given, $H_1$ should be $\Bbb{Z}^2$. If you're calculating the homology of the Klein bottle using the usual $\Delta$-complex structure, your second chain group should be $\Bbb{Z}^2$, rather than $\Bbb{Z}$. Idk if that's actually the space you're computing the homology of, but if you provide more details, perhaps we can be more helpful. $\endgroup$ – jgon Jun 12 at 22:34

The group is isomorphic to $\mathbb{Z} ^2$. Define the homomorphism $f:\mathbb{Z}^3 \rightarrow \mathbb{Z}^2$ by $(a,b,c)\rightarrow(b,2a-c)$. This map is onto and its kernel is $\langle (1,0,2) \rangle$. This is because you are in the kernel precisely when $b=0$ and $a,c$ satisfy $2a=c$, clearly any value of $a$ works and $c$ must be $2a$.

You can tell that the quotient can't be $\mathbb{Z}+\mathbb{Z}/2$ because if $(a,b,c)$ has order 2 in the quotient $b=0$ and we get $2(2a)=2b$ which implies $2a=b$ meaning it was already 0 in the quotient.

  • $\begingroup$ 2.2.28 in hatcher, calculating $H_1(X)$. This is just a part of the problem, but yeah you should look at it and type up an explanation for me hahaaaa $\endgroup$ – Mathematical Mushroom Jun 13 at 0:32
  • $\begingroup$ why does $2(2a)=2b$? did you mean $2(2a)=2c$? $\endgroup$ – Mathematical Mushroom Aug 1 at 16:11

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