[proofplan] We prove the assertion directly from the $\varepsilon$-$\delta$ definitions. [Uniform continuity](/page/Uniform%20Continuity) of $\Phi$ converts a desired $d$-error in $Z$ into a single $e$-scale in $Y$ that works for every pair of points in $Y$. [Uniform convergence](/page/Uniform%20Convergence) of $f_k$ to $f$ then guarantees that, eventually in $k$, the values $f_k(s)$ and $f(s)$ are within that $e$-scale for every $s\in S$, and composition with $\Phi$ gives the desired uniform estimate. [/proofplan] [step:Choose a uniform continuity scale for the target error] Fix $\varepsilon>0$. Since $\Phi:Y\to Z$ is uniformly continuous, there exists $\delta>0$ such that for all $y,y'\in Y$, \begin{align*} e(y,y')<\delta \implies d(\Phi(y),\Phi(y'))<\varepsilon. \end{align*} [guided] Fix a target error $\varepsilon>0$ in the codomain [metric space](/page/Metric%20Space) $(Z,d)$. To prove uniform convergence of the composed maps, we must eventually make \begin{align*} d((\Phi\circ f_k)(s),(\Phi\circ f)(s))<\varepsilon \end{align*} for every $s\in S$ at the same time. The hypothesis that $\Phi:Y\to Z$ is uniformly continuous is precisely what allows one scale in $Y$ to control this error uniformly over all possible inputs. Therefore there exists $\delta>0$ such that for every pair $y,y'\in Y$, \begin{align*} e(y,y')<\delta \implies d(\Phi(y),\Phi(y'))<\varepsilon. \end{align*} The important point is that $\delta$ depends only on $\varepsilon$ and on $\Phi$, not on the particular points $y,y'\in Y$ and not on the index $k$ or the point $s\in S$. [/guided] [/step] [step:Use uniform convergence to make every input pair $\delta$ close] Because $f_k:S\to Y$ converges uniformly to $f:S\to Y$ on $S$ with respect to $e$, the positive number $\delta$ chosen above gives an index $N\in\mathbb N$ such that for every $k\in\mathbb N$ with $k\ge N$ and every $s\in S$, \begin{align*} e(f_k(s),f(s))<\delta. \end{align*} [/step] [step:Apply the uniform continuity estimate pointwise and keep the index uniform] Let $k\in\mathbb N$ satisfy $k\ge N$, and let $s\in S$. The values $f_k(s)$ and $f(s)$ both belong to $Y$. Applying the implication from uniform continuity with $y=f_k(s)$ and $y'=f(s)$ gives \begin{align*} d(\Phi(f_k(s)),\Phi(f(s)))<\varepsilon. \end{align*} Since $(\Phi\circ f_k)(s)=\Phi(f_k(s))$ and $(\Phi\circ f)(s)=\Phi(f(s))$, this is \begin{align*} d((\Phi\circ f_k)(s),(\Phi\circ f)(s))<\varepsilon. \end{align*} The same index $N$ works for every $s\in S$, so $\Phi\circ f_k$ converges uniformly to $\Phi\circ f$ on $S$ with respect to $d$. [/step]