I am working on an exercise on injective modules:
Show that an $R$-module homomorphism $\alpha:A \to B$ is injective if the induced map Hom$_R(B,Q)\to$ Hom$_R(A,Q)$ is surjective for all injective modules $Q$. (Hint: $R$-Mod has enough injectives)
The fact that the map Hom$_R(B,Q)\to$ Hom$_R(A,Q)$ is surjective implies that every map $\varphi: A \to Q$ lifts to a map $\psi: B\to Q$, but I cannot see how this implies $\alpha$ is injective.
The hint refers to the fact that $R$-Mod has enough injectives. This is because for every $R$-module $M$, there exists an injective $I(M)$ with a map $M\to I(M)$. See this link.
You should embed the kernel $K$ into some injective module $Q$ and then apply $\operatorname{Hom}(-,Q)$ to the composition $$K \to A \to B.$$ You get a map which is both surjective and zero, hence $\operatorname{Hom}(K,Q)=0$. By our choice we have an embedding $K \subset Q$, so we deduce $K=0$.