Universal property of the quotient topology

Statement

Let $X$ be a topological space, let $\sim$ be an equivalence relation on $X$, and give $X / {\sim}$ the quotient topology, with quotient map $q \colon X \to X / {\sim}$. Let $Y$ be a topological space and let $g \colon X / {\sim} \to Y$ be a function. Then $g$ is continuous if and only if the composition $g \circ q \colon X \to Y$ is continuous.

Remarks

Proof of the converse: for $V$ open in $Y$, the set $q^{-1}(g^{-1}(V)) = (g \circ q)^{-1}(V)$ is open in $X$. Thus $g^{-1}(V)$ is open in the quotient. So each continuous map $f \colon X \to Y$ that is constant on each class gives a unique continuous map $g$ with $f = g \circ q$. For example, $t \mapsto (\cos 2\pi t, \sin 2\pi t)$ is constant on the classes of $\mathbb{R} / \mathbb{Z}$, so it gives a continuous map from $\mathbb{R} / \mathbb{Z}$ to the circle.

Explanations

L1 human

To define a continuous map on a quotient space, define a continuous map on the original space that takes the same value on glued points. The map then passes to the quotient.