[proofplan] We prove continuity by using the open-set definition. Given an arbitrary open subset of $Z$, we compute its preimage under $g \circ f$ as an iterated preimage: first under $g$, then under $f$. Continuity of $g$ makes the first preimage open in $Y$, and continuity of $f$ makes the second preimage open in $X$. [/proofplan] [step:Declare the composition as a map from $X$ to $Z$] Define the composition \begin{align*} g \circ f: X &\to Z \\ x &\mapsto g(f(x)). \end{align*} This is well-defined because $f(x) \in Y$ for every $x \in X$, and $g$ has domain $Y$. [/step] [step:Compute the preimage of an arbitrary open set in $Z$] Let $W \in \tau_Z$ be an arbitrary open subset of $Z$. By the definition of preimage and of composition, \begin{align*} (g \circ f)^{-1}(W) &= \{x \in X : (g \circ f)(x) \in W\} \\ &= \{x \in X : g(f(x)) \in W\} \\ &= \{x \in X : f(x) \in g^{-1}(W)\} \\ &= f^{-1}(g^{-1}(W)). \end{align*} [guided] We must prove that $g \circ f$ is continuous. By the [definition of continuity](/page/Continuity) for maps between topological spaces, it is enough to show that for every open subset $W$ of the codomain $Z$, the preimage $(g \circ f)^{-1}(W)$ is open in $X$. Fix an arbitrary open set $W \in \tau_Z$. We now compute the preimage of $W$ under the composed map. Using the definition of preimage and the definition of composition, \begin{align*} (g \circ f)^{-1}(W) &= \{x \in X : (g \circ f)(x) \in W\} \\ &= \{x \in X : g(f(x)) \in W\}. \end{align*} The condition $g(f(x)) \in W$ is equivalent to saying that $f(x)$ lies in the preimage of $W$ under $g$. Since \begin{align*} g^{-1}(W) = \{y \in Y : g(y) \in W\}, \end{align*} we obtain \begin{align*} \{x \in X : g(f(x)) \in W\} &= \{x \in X : f(x) \in g^{-1}(W)\} \\ &= f^{-1}(g^{-1}(W)). \end{align*} Thus the preimage of $W$ under the composition is exactly the preimage under $f$ of the preimage under $g$. [/guided] [/step] [step:Use continuity of $g$ and then continuity of $f$] Since $g: Y \to Z$ is continuous and $W \in \tau_Z$, the set $g^{-1}(W)$ belongs to $\tau_Y$. Since $f: X \to Y$ is continuous and $g^{-1}(W) \in \tau_Y$, the set $f^{-1}(g^{-1}(W))$ belongs to $\tau_X$. Therefore \begin{align*} (g \circ f)^{-1}(W) \in \tau_X. \end{align*} [guided] We now use the two continuity hypotheses in the correct order. First, $W$ is open in $Z$, meaning $W \in \tau_Z$. Since $g: Y \to Z$ is continuous, the preimage of every open subset of $Z$ under $g$ is open in $Y$. Applying this to $W$, we get \begin{align*} g^{-1}(W) \in \tau_Y. \end{align*} Next, since $f: X \to Y$ is continuous, the preimage under $f$ of every open subset of $Y$ is open in $X$. The set $g^{-1}(W)$ is an open subset of $Y$, so applying continuity of $f$ gives \begin{align*} f^{-1}(g^{-1}(W)) \in \tau_X. \end{align*} From the preimage identity computed above, \begin{align*} (g \circ f)^{-1}(W) = f^{-1}(g^{-1}(W)), \end{align*} and hence \begin{align*} (g \circ f)^{-1}(W) \in \tau_X. \end{align*} [/guided] [/step] [step:Conclude that the composition is continuous] Because $W \in \tau_Z$ was arbitrary, the preimage of every open subset of $Z$ under $g \circ f$ is open in $X$. By the open-set definition of continuity, $g \circ f: X \to Z$ is continuous. [/step]