# How to prove that some set has the quotient topology for a function?

Let $f: X \to Y$ be a continuous function between topological spaces. Let $S$ be a set and $g: Y \to S$ a function. Assume that $g \circ f$ is surjective, and that $S$ has the quotient topology for $g \circ f$. Assume that $g$ is continuous. How does one show that $S$ has the quotient topology for $g$?

The definition of the quotient topology for $g$ on $S$ is (I think): $T = \{ U \subset S | g^{-1}(U) \space \text{open in} Y \}$.

This was an question for a previous exam for our topology course. I don't really know how to prove this.

• If $g \circ f$ is surjective, then $g$ is surjective. This, in combination with the continuity of the composition and $g$ gives your answer. Write out what the preimage of an open set through $g \circ f$ is, and look at what it means for $g$ specifically. – Andy Jun 3 '13 at 10:38

Assume that $h:S\to T$ is a set map between the spaces $S$ and $T$ such that $h\circ g$ is continuous. Then $h\circ g\circ f$ is also continuous, and since $S$ has the quotient topology with respect to $g\circ f$, this implies that $h$ is continuous. Hence $S$ has the universal property characterizing the quotient topology.