Proof that the composition of continuous maps is continuous

Proof

We check the definition of continuity with preimages of open sets.

1. Let $X$, $Y$ and $Z$ be topological spaces, and let $f \colon X \to Y$ and $g \colon Y \to Z$ be continuous.
2. Let $W$ be an open set of $Z$. We show that the preimage $(g \circ f)^{-1}(W)$ is open in $X$.
3. Let $x \in X$. By the definition of the composition, $x \in (g \circ f)^{-1}(W)$ holds if and only if $g(f(x)) \in W$.
4. This holds if and only if $f(x) \in g^{-1}(W)$, which holds if and only if $x \in f^{-1}(g^{-1}(W))$.
5. By steps 3 and 4, $(g \circ f)^{-1}(W) = f^{-1}(g^{-1}(W))$.
6. The map $g$ is continuous and $W$ is open in $Z$, so $g^{-1}(W)$ is open in $Y$.
7. The map $f$ is continuous and $g^{-1}(W)$ is open in $Y$, so $f^{-1}(g^{-1}(W))$ is open in $X$.
8. By steps 5 and 7, $(g \circ f)^{-1}(W)$ is open in $X$ for each open set $W$ of $Z$. So $g \circ f$ is continuous.