Proof of the universal property of the quotient topology

Proof

We prove the two directions separately. Let $\tau$ be the topology of $X$.

1. The quotient map $q$ is continuous. Indeed, if $V$ is open in $X / {\sim}$, then $q^{-1}(V) \in \tau$ by the definition of the quotient topology.
2. Let $g$ be continuous. Then $g \circ q$ is a composition of continuous maps, so it is continuous.
3. Conversely, let $g \circ q$ be continuous, and let $V$ be open in $Y$.
4. By the definition of the composition, $x \in (g \circ q)^{-1}(V)$ if and only if $g(q(x)) \in V$. This holds if and only if $q(x) \in g^{-1}(V)$, that is, $x \in q^{-1}(g^{-1}(V))$.
5. So the preimages agree: $q^{-1}(g^{-1}(V)) = (g \circ q)^{-1}(V)$. This set is in $\tau$ because $g \circ q$ is continuous.
6. By the definition of the quotient topology, $g^{-1}(V)$ is open in $X / {\sim}$.
7. So the preimage under $g$ of each open set of $Y$ is open, and $g$ is continuous.