Does $ 0 \rightarrow A \rightarrow B \rightarrow A \rightarrow 0$ being exact implies $A \cong B $

Question

As in the title I would like to know if $ 0 \rightarrow A \rightarrow B \rightarrow A \rightarrow 0$ being exact implies $A \cong B $. Is it true at least when $A=0 $? Thank you in advance

Answer 1

When $A=0$, then you have an exact sequence $0\to B\to 0$, and thus $B=0$.

In general this is false, for example, the sequence $0\to \mathbb Z\to\mathbb Z\oplus \mathbb Z\to \mathbb Z\to 0$, with the first map being inclusion $\mathbb Z\hookrightarrow\mathbb Z\oplus\{0\}$ and the second being projection onto the second factor, is exact, but clearly $\mathbb Z$ is not isomorphic to $\mathbb Z\oplus \mathbb Z$.