Lifting a homomorphism between modules in exact sequences > Let $0\to A\to P\to B\to 0$ and $0\to C\to Q \to D\to 0$ be two exact sequences of $R$ modules, with $Q$ injective, $\operatorname{Hom}(B,D)=0$, and let there be a homomorphism $\alpha:A\to C$. I'm trying to show that $\alpha$ lifts to a homomorphism $a^*:P\to C$. The only way I can see of making use of the injectivity of $Q$ is by applying the functor $\operatorname{Hom}(\cdot,Q)$ to both sequences, and examining the resulting exact sequences. From here however, I cannot see how to use the fact that $\operatorname{Hom}(B,D)=0$. I was thinking of using the _five lemma_ , but do not seem to be in the possession of enough mono- and epi- morphisms. Any hints on how to proceed would be appreciated.

$$\require{AMScd} \begin{CD} 0 @>>> A @>\alpha>> P @>\beta>> B @>>> 0 \\\ @. @VfVV \\\ 0 @>>> C @>\gamma>> Q @>\delta>> D @>>> 0 \end{CD} $$ Since $Q$ is injective, there exists $g\colon P\to Q$ such that $g\alpha=\gamma f$: \begin{CD} 0 @>>> A @>\alpha>> P @>\beta>> B @>>> 0 \\\ @. @VfVV @VgVV \\\ 0 @>>> C @>\gamma>> Q @>\delta>> D @>>> 0 \end{CD} Now $\delta g\alpha=\delta\gamma f=0$, so there exists $h\colon B\to D$ such that $h\beta=\delta g$ (homomorphism theorem): \begin{CD} 0 @>>> A @>\alpha>> P @>\beta>> B @>>> 0 \\\ @. @VfVV @VgVV @VhVV \\\ 0 @>>> C @>\gamma>> Q @>\delta>> D @>>> 0 \end{CD} By assumption, $h=0$, so also $\delta g=0$, which implies the image of $g$ is contained in the kernel of $\delta$, which is the same as the image of $\gamma$.