Prove the dual of Schanuel's Lemma. Given exact sequences \begin{equation}0 \rightarrow M \stackrel{i}{\rightarrow} E \stackrel{p}{\rightarrow} Q \rightarrow 0\end{equation} and \begin{equation}0 \rightarrow M \stackrel{i'}{\rightarrow} E' \stackrel{p'}{\rightarrow} Q' \rightarrow 0\end{equation} where $E$ and $E'$ are injective, then there is an isomorphism $Q\oplus E'\cong Q'\oplus E$.
My attempt. If there is a following commute diagram. Then just imitate the proof of Schanuel's Lemma.
Since $E'$ is injective, so there exists a $f$ such that the left square commute. But I don't know how to get a $g$ such that the right square commute. Can anyone help me?
As $p'\circ f\circ i=p'\circ i'=0$ then $p'\circ f$ factors (uniquely) through $p$, as $p$ is the cokernel of $i$. That is, there is $g$ with $p'\circ f=g\circ p$.