$f: A \to B$ continuous and $g \circ f : A \to C$ continuous implies $g : B \to C$ continuous?

Question

I have a function $f$ defined on a compact subset of $K \subset \mathbb{R}^n$ which is maps $K$ continuous and in a surjective manner to a metric space $(B,d)$. Assume $g: B \to C$ is a function from $B$ to a metric space $C$ such that $g \circ f$ is continuous. Can we conclude that $g$ is continuous as well?

Answer 1

Yes, you can.

Taking complements it suffices to show that $g^{-1}(F)$ is closed for every closed $F\subset C$. Since $f$ is surjective, it follows from basic results about images and preimages that $$g^{-1}(F)=f(f^{-1}(g^{-1}(F))=f((g\circ f)^{-1}(F)).$$ Since $g\circ f$ is continuous, the set $(g\circ f)^{-1}(F)$ is a closed subset of the compact space $K$, hence compact, and since $f$ is continuous, its image under $f$ is compact. Thus $g^{-1}(F)$ is closed.