$f:X\to Y$ continuous, $X$ compact, $g:Y\to Z$ such that $g\circ f$ is continuous $\implies$ $g$ is continuous

Question

Let $f:X\to Y$ be continuous map onto $Y$, and let $X$ be compact. Also $g:Y\to Z$ is such that $g\circ f$ is continuous. I have to show that $g$ is continuous.

Facts obvious from the assumptions:

• As continuous image of compact set is compact, $Y$ and $g(Y)\subseteq Z$ are compact.
• Let $B$ be an open subset of $g(Y)$, then $f^{-1}(g^{-1}(B))$ is an open subset of $X$.

How do I show that $g^{-1}(B)$ is open in $Y$? I have to use the property of compactness somewhere, but not sure how.

Answer 1

I will assume $Y$ is additionally Hausdorff. Otherwise I don't think this is true or at least I don't see how it works in non-Hausdorff case.

In this situation since $X$ is compact then $f:X\to Y$ is a closed map by the closed map lemma (the assumption about $Y$ being Hausdorff kicks in here). And since $f$ is onto then $f$ is a quotient map, meaning $U\subseteq Y$ is open if and only if $f^{-1}(U)$ is open in $X$.

With that take an open subset $V\subseteq Z$. Then $(g\circ f)^{-1}(V)$ is open in $X$ since we assumed $g\circ f$ is continuous, i.e. $f^{-1}(g^{-1}(V))$ is open in $X$. Since $f$ is a quotient map then this is if and only if $g^{-1}(V)$ is open in $Y$ and thus $g$ is continuous.

Answer 2

This need not hold unless you put some separation condition (e.g. Hausdorff separation, as done above) on $Y$. I will construct a counterexample.

Let $X$ be the set $\{0,1\}$ with the discrete topology. Let $Y,Z$ be the set $\{0,1\}$ equipped with the Alexandroff topology given by the order relation $0 \leq 1$. These spaces are compact and T0-separable, but $Y$ is not Hausdorff.

The inclusion function $f: X \rightarrow Y$ is continuous and onto. The function $g: Y \rightarrow Z$ given by $g(y) = 1 - y$ is not continuous since $g(0) \not \leq g(1)$. However, $g \circ f: X \rightarrow Z$ is continuous since every function from a discrete space to another topological space is continuous.