[proofplan] We prove continuity of $f \circ g$ at $a$ directly from the relative $\varepsilon$-$\delta$ definition. Given an error tolerance for $f \circ g$, continuity of $f$ at $g(a)$ gives a tolerance for the input of $f$. Continuity of $g$ at $a$ then gives a neighbourhood of $a$ in $E$ on which $g(x)$ lies within that input tolerance, allowing the two estimates to compose. [/proofplan] [step:Declare the composition as a function on $E$] Since $g: E \to F$ and $f: F \to \mathbb{R}$, the composition is the function \begin{align*} f \circ g: E &\to \mathbb{R} \end{align*} defined by $(f \circ g)(x)=f(g(x))$ for every $x \in E$. In particular, because $a \in E$, we have $g(a) \in F$, so the hypothesis that $f$ is continuous at $g(a)$ is well-typed. [/step] [step:Choose the tolerance for $g$ from continuity of $f$] Let $\varepsilon > 0$ be given. Since $f: F \to \mathbb{R}$ is continuous at $g(a)$ relative to $F$, there exists $\eta > 0$ such that for every $y \in F$, \begin{align*} |y-g(a)| < \eta \implies |f(y)-f(g(a))| < \varepsilon. \end{align*} [guided] We begin with the quantity that must be made small for continuity of $f \circ g$ at $a$, namely \begin{align*} |(f \circ g)(x)-(f \circ g)(a)|. \end{align*} By the definition of composition, this is \begin{align*} |f(g(x))-f(g(a))|. \end{align*} Thus the relevant continuity hypothesis is the continuity of $f$ at the point $g(a) \in F$. Let $\varepsilon > 0$ be given. The relative continuity of $f: F \to \mathbb{R}$ at $g(a)$ says that there exists an input tolerance $\eta > 0$ such that every point $y \in F$ satisfying $|y-g(a)|<\eta$ also satisfies \begin{align*} |f(y)-f(g(a))| < \varepsilon. \end{align*} The restriction $y \in F$ matters because $f$ is only defined on $F$, not necessarily on all of $\mathbb{R}$. [/guided] [/step] [step:Use continuity of $g$ to force $g(x)$ into the required tolerance] Since $g: E \to F$ is continuous at $a$ relative to $E$, applied with the positive number $\eta$, there exists $\delta > 0$ such that for every $x \in E$, \begin{align*} |x-a| < \delta \implies |g(x)-g(a)| < \eta. \end{align*} [/step] [step:Combine the two implications to prove continuity of the composition] Let $x \in E$ satisfy $|x-a|<\delta$. By the choice of $\delta$, \begin{align*} |g(x)-g(a)| < \eta. \end{align*} Since $g(x) \in F$, the choice of $\eta$ applies with $y=g(x)$ and gives \begin{align*} |f(g(x))-f(g(a))| < \varepsilon. \end{align*} Using the definition of the composition at $x$ and at $a$, this is exactly \begin{align*} |(f \circ g)(x)-(f \circ g)(a)| < \varepsilon. \end{align*} Therefore, for every $\varepsilon>0$ there exists $\delta>0$ such that every $x \in E$ with $|x-a|<\delta$ satisfies the displayed inequality. Hence $f \circ g$ is continuous at $a$ relative to $E$. [/step]