Surjectivity of induced map via hom functor implies injectivity

[Kreizhn]

Homework Statement

Let R be an arbitrary ring, B and B' be left R-modules, and $i: B' \to B$ be an R-module morphism. Show that if the induced map $i^*: \operatorname{Hom}_R(B,M) \to \operatorname{Hom}(B',M)$ is surjective for every R-module M, then $i: B' \to B$ is injective.

The Attempt at a Solution

The maps all seem to go the wrong way to use the categorical definition of kernels, so I fear that I must be much trickier about the application and exploit the module structure quite specifically. This would suggest an intelligent choice of M and a morphism $B' \to M$ in order to apply the hypothesis.

In my mind, the only obvious candidate is the projection map $\pi: B' \to B'/\ker i$. The hypothesis would then suggest that there exists $\hat \pi: B \to B'/\ker i$ such that $\hat \pi \circ i = \pi$. However, nothing useful seems to come of this. Any ideas?

 

[micromass]

What if you take $M=B^\prime$ and take the identity morphism in $Hom(B^\prime,B^\prime)$?

 

[Kreizhn]

Ah yes excellent. I figured that out today, and could have saved myself a lot of time if I had just looked here first.