Continuity of the Composition of Continuous Maps on Topo. Spaces

On the Summary of Equivalent Statements Regarding Continuous Maps on Topological Spaces page we summarizes all of the equivalent states for a map $f : X \to Y$ on topological spaces to be continuous on all of $X$. We will now look at a nice theorem which tells us that the composition of continuous maps on topological spaces will be continuous.

Theorem 1: Let $X$, $Y$, and $Z$ be topological spaces and let $f : X \to Y$ be continuous on all of $X$ and $g : Y \to Z$ be continuous on all of $Y$. Then $g \circ f : X \to Z$ is continuous of all of $X$.

• Proof: Since $f$ is continuous on all of $X$ we have that for every open set $V$ in $Y$ that $f^{-1}(V)$ is open in $X$.

• Similarly, since $g$ is continuous on all of $Y$ we have that for every open set $W$ in $Z$ that $g^{-1}(W)$ is open in $Y$.

• Consider the composite function $g \circ f : X \to Z$. Let $W$ be an open set in $Z$. Then:

(1)

\begin{align} \quad (g \circ f)^{-1} (W) = f^{-1}(g^{-1}(W)) \end{align}

• But $g^{-1}(W)$ is open in $Y$, so let $V = g^{-1}(W)$. Then:

(2)

\begin{align} \quad (g \circ f)^{-1}(W) = f^{-1}(V) \end{align}

• But $f^{-1}(V)$ is open in $X$. Therefore $(g \circ f)^{-1}(W)$ is open in $X$. Since $W$ is an arbitrary open set in $Z$ we see that for every open set $W$ in $Z$ that then $(g \circ f)^{-1} (W)$ is open. Hence $g \circ f$ is continuous on all of $X$. $\blacksquare$