= Solution
Take $B=\mathbb Z$, $N=\mathbb Z/2\mathbb Z$, and the <short exact sequence>
$$
0\longrightarrow\mathbb Z\xrightarrow{\times2}\mathbb Z\longrightarrow\mathbb Z/2\mathbb Z\longrightarrow0.
$$
Multiplication 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 gives
$$
0\longrightarrow0\longrightarrow0\longrightarrow\mathbb Z/2\mathbb Z\longrightarrow0.
$$
\b[This 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 $N\to\mathbb Z$.
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