I haven't had time to think about this yet, but it seems interesting and not-entirely trivial, so I'll post it here.
Given a continuous mapping $f : A \rightarrow B$ between topological spaces, lets say that $f$ is simple if and only if the corresponding map $\hat{f} : A \rightarrow f(A)$ is a quotient mapping.
Question. If $f : A \rightarrow B$ and $g : B \rightarrow C$ are simple continuous mappings between topological spaces, is $g \circ f$ also necessarily simple?
The question really boils down to the following: If $q:X \to Y$ is a quotient map, and $A ⊆ X$, is $q\restriction{A}: A \to q[A]$ also quotient?
The answer is no in a rather strong way. Take $q:ℝ \oplus ℝ \to ℝ$ gluing the two copies of $ℝ$ in the corresponding points. The map $q$ is really nice – not only quotient but also closed and open at the same time. On the other hand, $q\restriction{ℚ \oplus (ℝ \setminus ℚ)}$ is a continuous bijection, but clearly not a homeomorphism and so not a quotient.
There is a notion of hereditarily quotient map (2.4.F in Engelking) – every co-restriction $q\restriction{q^{-1}[B]}: q^{-1}[B] \to B$ should be quotient. Every closed quotient or open quotient map is hereditarily quotient, but not every quotient map is hereditarily quotient.
A minimalistic counterexample: Let $X$ be the set $\{0, 1, a_0, a_1\}$ with the topology generated by the set $\{1, a_1\}$. Let $q: X \to Y = \{0, 1, a\}$ be the quotient map gluing the points $a_0$, $a_1$ together. The space $Y$ is indiscrete, and so the co-restriction $q\restriction{q^{-1}[\{0, 1\}]}: \{0, 1\} \to \{0, 1\}$ is not quotient.
Another example. Not every quotient of a Fréchet–Urysohn space is Fréchet–Urysohn. In fact, these are exactly sequential spaces. On the other hand, hereditary quotients preserve being Fréchet–Urysohn. So for example the quotient map from the topological sum of $ω+1$ copies of the convergent sequence $ω+1$ onto the Arens' space is not hereditarily quotient. Here the source space is countable metrizable and the target space is countable zero-dimensional.