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From this problem here, it says that any counter example can not be finitely generated Abelian groups. But I can not find the mistake in the following counter example, I must missed something obvious, $$0\rightarrow \mathbb{Z} \xrightarrow{ z\times2 \oplus z\text{ mod }2} \mathbb{Z}\oplus \mathbb{Z}/2 \xrightarrow{ \text{ mod }2\circ\pi_\mathbb{Z} } \mathbb{Z}/2 \rightarrow 0.$$

The first map is injective, second map is surjective, and the image of first map is the kernel of the second map.

And it fails to be exact with the same reason for $0\rightarrow \mathbb{Z}\xrightarrow{z\times 2} \mathbb{Z} \xrightarrow {(\text{mod }2)}\mathbb{Z}/2 \rightarrow 0$, there is no non-zero homomorphism from $\mathbb{Z}/2$ to $\mathbb{Z}$.

Thank you.

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1 Answer 1

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I presume your first map is $a\mapsto(2a,\overline a)$. I cannot read what your second map is. But the image $I$ of the first map has index $4$ in $\Bbb Z\oplus\Bbb Z/2$, so it cannot be the kernel of your second map. Observe that $(0,\overline 0)$, $(0,\overline 1)$, $(1,\overline 0)$ and $(1,\overline 1)$ are in distinct cosets of $I$.

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  • $\begingroup$ The second map is projection onto the first component (which is $\mathbb{Z}$) then compose mod 2. $\endgroup$
    – Xiao
    Commented Sep 24, 2017 at 18:42
  • $\begingroup$ So it is $(a,\overline b)\mapsto\overline a$. Then $(0,\overline 1)$ is in its kernel but not in the image of the first map. @Xiao $\endgroup$ Commented Sep 24, 2017 at 18:44
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    $\begingroup$ I see, thank you very much! $\endgroup$
    – Xiao
    Commented Sep 24, 2017 at 18:45

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