Here is Munkres' way of phrasing the universal property of quotient maps:
Let $p : X \to Y$ be a quotient map. Let $Z$ be a space and let $g : X \to Z$ be a map that is constant on each set $p^{-1}(\{y\})$, for $y \in Y$. Then $g$ induces a map $f : Y \to Z$ such that $f \circ p = g$. The induced map $f$ is continuous if and only if $g$ is continuous; $f$ is a quotient map if and only if $g$ is a quotient map.
Notice that it omits the very important word "unique", which other sources include (see this and this). However, these other sources also stipulate that $g$ (using my notation) is continuous. My question is, does uniqueness follow if $g$ is also assumed to be continuous, or does uniqueness of $f$ just follow from $f \circ p = g$, so that if $h : Y \to Z$ is some other map such that $h \circ p = g$, it follows that $h = f$, independently of whether $g$ or $f$ is continuous?
I ask because this subtle point has come up in a problem I'm working (trying to show that $[0,1]/ \{0,1\}$ is homeomorphic to $S^1$).
$p$ is onto in Munkres' definition. It follows from this and $f \circ p = g$ that $f$ is unique: take $y \in Y$. $y = p(x)$ for some $x \in X$. By the commutativity condition we get that $f(y)= f(p(x)) = g(x)$ is forced. It's given that $g$ is constant on the fibres of $p$ so the choice of $x$ does not matter.