My text book asks me to come up with an example when $n \circ m$ is monic, thoguht $n$ itself is not. Consider following proof:
Given: $(n \circ m) \circ f = (n \circ m) \circ g \implies f = g$ which is the same as $n \circ (m \circ f) = n \circ (m \circ g) \implies f = g$ (associativity, thank you!). Lemma: $\forall k: f = g \implies k \circ f = k \circ g$. By the given lemma $f = g \implies (m \circ f) = (m \circ g)$, which is $$n \circ (m \circ f) = n \circ (m \circ g) \implies (m \circ f) = (m \circ g) \implies f = g$$
Clearly, both $n, m$ are monics, thus there exists no case when $m$ is monic and yet $n$ isn't.
Am I right or my proof is wrong?
In SET, we have that the includion $m\colon\{a\}\to\{a,b\}$ is monic, the constant map $n\colon \{a,b\}\to\{a\}$ is not monic, and yet $n\circ m$ is monic. For other categories it is often (e.g., when finite direct sums exist) just as easy to come up with similar counterexamples.