Consider an exact sequence of abelian groups $$ 0 \to A \to \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \to B \to 0, $$ where we make no assumption on the map $\mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$. Is it possible to determine $A$ and $B$ up to isomorphism of abelian groups from this data?
2026-03-25 23:56:32.1774482992
An exact sequence of abelian groups
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No, $A$ and $B$ depend on the map $\phi:\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}$.
If $\phi$ is trivial then $(A,B)=(\mathbb{Z}\oplus\mathbb{Z},\mathbb{Z})$ and if for example $\phi$ is the projection onto one copy of $\mathbb{Z}$ then $(A,B)=(\mathbb{Z},1)$ (where $1$ is the trivial group).
These are not the only possibilities, we can take any homomorphism $\phi$, then $A\cong\ker(\phi)$ and $B\cong\mathbb{Z}/{\rm im}(\phi)$