 Solution

Take $B=\mathbb{Z}$, $N=\mathbb{Z}/2\mathbb{Z}$, and the short exact sequenceMultiplication by two is injective, and the reduction map has image all of $\mathbb{Z}/2\mathbb{Z}$ and kernel $2\mathbb{Z}$, proving exactness. Any module homomorphism from $\mathbb{Z}/2\mathbb{Z}$ to $\mathbb{Z}$ is zero: its generator must map to an integer killed by two. On the other hand, a homomorphism from $\mathbb{Z}/2\mathbb{Z}$ to itself is determined arbitrarily by the image of its generator, so its Hom group is $\mathbb{Z}/2\mathbb{Z}$. Applying the Hom functor therefore givesThis sequence is not exact at $\mathbb{Z}/2\mathbb{Z}$: the incoming image is zero while the outgoing kernel is the whole nonzero group. In particular, the identity of $N$ cannot lift through the reduction map to a map $N\to \mathbb{Z}$.