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?
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.