The following is quoted from https://en.wikipedia.org/wiki/Quotient_space_(topology)
Quotient maps q : X → Y are characterized among surjective maps by the following property: if Z is any topological space and f : Y → Z is any function, then f is continuous if and only if f ∘ q is continuous.
Assuming that $q$ is a quotient map, I can prove the characterizing property using the definition of a quotient map, namely, $q^{-1}U$ is open if and only if $U$ is open. However, I could not see how to prove the converse: how does the property imply that $q$ is quotient?
Best Answer
Suppose $q: X \to Y$ obeys the "composition property". Suppose $U$ is a subset of $Y$ such that $q^{-1}[U]$ is open. Define $g: Y \to \{0,1\}$, where $\{0,1\}$ has the Sierpinski topology $\{\emptyset,\{0\},\{0,1\}\}$, by $g(x) = 0$ when $x \in U$ and $g(x)=1$ for $x \notin U$.
Then $g \circ q: X \to \{0,1\}$ sends $q^{-1}[U]$ to $0$ and all other points of $X$ to $1$ and because we assumed $q^{-1}[U]$ is open and as $\{0,1\}$ has the specified topology, $g \circ q$ is continuous. So by the "composition property", $g$ must be continuous, so $g^{-1}[\{0\}]=U$ must be open.
So we have show that the topology of $Y$ contains all $U$ such that $q^{-1}[U]$ is open and from this it follows that $q$ is a quotient map.