I have an exact sequence ending with \begin{equation*} \cdots \rightarrow A \xrightarrow{f_1} \mathbb{Z} \xrightarrow{f_2} \mathbb{Z} \oplus \mathbb{Z} \xrightarrow{f_3} \mathbb{Z} \rightarrow (0) \end{equation*}
Can I say that $Im(f_1) = (0)$? My argument goes like this: we know $f_3$ is surjective, so since $Ker(f_3) = Im(f_2)$, \begin{equation*} \mathbb{Z} \oplus \mathbb{Z} / Im(f_2) \cong \mathbb{Z} \end{equation*} But $Im(f_2)$ is a subgroup of a free abelian group of rank $2$, so it is itself a free abelian group, and the rank must be $1$ because of this quotient relation, hence $Im(f_2) \cong \mathbb{Z}$.
Then, we obtain, since $Ker(f_2) = Im(f_1)$, \begin{equation*} \mathbb{Z} / Im(f_1) \cong Im(f_2) \cong \mathbb{Z} \end{equation*}
So it seems trivial to me that must have $Im(f_1) = (0)$. Is that correct or am I missing something? Thanks in advance.