Why we can identify $A$ with its image $f(A)$ in a short exact sequence?

Question

Consider the short exact sequence $0 \longrightarrow A \stackrel{f}{\longrightarrow} B \stackrel{g}{\longrightarrow} C \longrightarrow 0$ the book says that we can identify $A$ with its image $f(A)$ but i dont understand why can you help me to understand?

Answer 1

For a short exact sequence of the form mentioned, $f$ is injective, $g$ is surjective, and $Im(f)=Ker(g)$. Since $f$ is injective, considering $f$ as a map from $f:A\to Im(f)$, one can say $f$ is also surjective, and hence $A$ is in one-one bijection with $Im(f)=f(A)$.