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Suppose we have the following short exact sequence of abelian groups: $$ 0 \rightarrow A \xrightarrow{a} B \xrightarrow{b} \mathbb{Z}/2 \rightarrow 0. $$

Under which conditions do we have that $b^{-1}(1)$ is isomorphic, as a set, to $A$?

I suspect this happens when the sequence splits (so then we have $B \cong A \oplus \mathbb{Z}/2$, and the preimage of $1$ under $b$ is just a copy of $A$), but I am having some trouble showing this rigorously. Also, is it true that this sequence always splits?

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    $\begingroup$ I suppose by isomorphism $b^{-1}(1)\cong A$ you mean that of abelian groups. But why should the coset $b^{-1}(1)$ be a group? $\endgroup$
    – Phil. Z
    Nov 14, 2017 at 9:18
  • $\begingroup$ @Phil.Z actually no, it's sufficient for me to show they are isomorphic as sets, I'll add this clarification $\endgroup$
    – Vitaly B
    Nov 14, 2017 at 9:20
  • $\begingroup$ Well, then the answer is always, since $b^{-1}(1)$ is a coset. $\endgroup$
    – Phil. Z
    Nov 14, 2017 at 9:24

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If all you want is a bijection between $b^{-1}(1)$ and $A$, then you always have one, as $b^{-1}(1)$ is the coset of $\ker(b)$ not containing zero, and $\ker(b)$ is isomorphic to $A$ by exactness.

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  • $\begingroup$ I see, I was overlooking the fact we should view $b^{-1}$ as a coset, so it has the same cardinality as $\ker(b) \cong A$. Thanks! $\endgroup$
    – Vitaly B
    Nov 14, 2017 at 9:31

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