The following is from Problem 13 of Chapter 3 in Awodey's Category Theory.
Let $f,g:M\to N$ be fixed monoid homomorphisms. Define two equivalence relations on the monoid $N$ as follows:
(1) $n\sim n'$ $\iff$ for all monoids $X$ and homomorphisms $h:N\to X$, one has $h\circ f=h\circ g$ implies $h(n)=h(n')$,
(2) the intersection of all equivalence relations $\sim$ on $N$ satisfying $f(m)\sim g(m)$ for all $m\in M$ as well as $$n\sim n' \text{ and } m\sim m'\implies n\cdot m\sim n'\cdot m'.$$
Show that these equivalence relations on $N$ agree.
For clarity let's call the equivalence relation given in (1) $R$ and the one given in (2) $S$. It was pretty easy to show $S\subset R$ since $R$ equates elements $f(m)$ and $g(m)$ and $R$ is closed under multiplication in the sense given in (2). However, I have no idea how to show the reverse containment. Maybe I'm just being obtuse, but regardless help would be appreciated.
To show $R\subseteq S$:
Assume that $nRn'$, i.e., that $h(n)=h(n')$ for all monoid homomorphisms $h$ with domain $N$ and $h\circ f=h\circ g$.
We want to show that $n\sim n'$ for all equivalence relations $\sim$ on $N$ satisfying
$$\tag1\forall m\in M\colon\ f(m)\sim g(m)$$ $$\tag2\forall a,a',b,b'\in N\colon \ a\sim a'\land b\sim b'\to ab\sim a'b'.$$
So let $\sim$ be any equivalence relation on $N$ with properties $(1)$ and $(2)$. Write $[x]:=\{\,y\in N\mid y\sim x\,\}$ for the equivalence class of $x\in N$ and let $X=\{\,[x]\mid x\in N\,\}$. We can define a multiplication on $X$ per $[x]\cdot[y]:=[xy]$ (that this is well-defined follows from $(2)$). This turns $X$ into a monoid. Note that $h\colon N\to X$, $x\mapsto [x]$ is a monoid homomorphism. We have $h\circ f=h\circ g$ by $(1)$. We conclude that $h(n)=h(n')$, in other words $[n]=[n']$, or: $$\tag3n\sim n'.$$ So $(3)$ holds for every equivalence relation on $N$ with $(1)$ and $(2)$, it follows that $nSn'$.