S.E.S given by: $0 \rightarrow \mathbb{Z} \stackrel{f}{\rightarrow} \mathbb{Z}^3 \stackrel{h}{\rightarrow} H_1(X) \rightarrow 0$

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

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!

• 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. – 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.
• 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 – Mathematical Mushroom Jun 13 at 0:32
• why does $2(2a)=2b$? did you mean $2(2a)=2c$? – Mathematical Mushroom Aug 1 at 16:11