Counterexample: surjective module homomorphism in an exact sequence

Question

I need to give an example to show that the exactness of $0 \to M_1 \to M_2 \to M_3 \to 0$ need not imply that the map $\operatorname{Hom}_R(N,M_2) \to \operatorname{Hom}_R(N,M_3)$ is surjective, where the latter is defined as $h \mapsto g \circ h$. I really suck at coming up with counter-examples; I don't even know where to start. I have tried the exact sequence: $$ 0 \to 2\mathbf{Z} \to \mathbf{Z} \to \mathbf{Z}/2\mathbf{Z} \to 0, $$ over the ring $R = \mathbf{Z}$. However, it seems to be true that every element in $\operatorname{Hom}_\mathbf{Z}(\mathbf{Z},\mathbf{Z}/2\mathbf{Z})$ can be written as a composition of an $R$-homomorphism $\mathbf{Z} \to \mathbf{Z}$ and the projection map. Does someone have some suggestions for counterexamples? I don't have any toolbox of counterexamples available yet in this subject, so I don't know where to start.

Answer 1

In general, $\text{Hom}(N, -)$ is a left exact functor for any $N$. The problem you're encountering in constructing counter-examples is that $\text{Hom}_R(N, -)$ is right exact precisely when $N$ is a projective $R$-module. In your case, choosing $N = \mathbb{Z}$ was not going to get you anywhere, because $\text{Hom}_\mathbb{Z}(\mathbb{Z}, -)$ is exact.

However, you are very close to a good counterexample. Try taking $N = \mathbb{Z}/2\mathbb{Z}$ instead (note $\mathbb{Z}/2\mathbb{Z}$ is not even flat as a $\mathbb{Z}$-module).

Now, can you compute $\text{Hom}(\mathbb{Z}/n\mathbb{Z}, \mathbb{Z})$?