TITLE: Show $F:C[0,1] \rightarrow C[0,1]$ defined by $F(f) = g \circ f$ is continuous.
QUESTION [5 upvotes]: Define $F:C[0,1] \rightarrow C[0,1]$ by $F(f) = g \circ f$, where $g:\Bbb R \rightarrow \Bbb R$ is continuous.
I am struggling to show that F is continuous. We use the supremum norm: 
$$ ||f|| = sup_{x\in[0,1]}|f(x)| $$
Let U be an open set in $C[0,1]$ (in the range of F). If $h \in U$, then $h$ is of the form $h = g \circ f$. The composition of continuous functions is continuous, so $h = g \circ f$ is continuous. Since $U$ open, $\exists \epsilon \gt 0$ such that $B_{\epsilon}(h) \subseteq U$.
Now, $F^{-1}(U) = \{f:g \circ f \in U\}$ Thus since $U$ is open, $F^{-1}(U)$ is open (how can I assert this)?
Any hints are greatly appreciated!

REPLY [2 votes]: Let's see that $F$ is continuous at some $f_0$. Let $\varepsilon > 0$. By uniform continuity of $g$ in $B_{1+\|f_0\|}(0)$ we have $\delta > 0$ such that:
$$
|x - y| < \delta \Rightarrow |g(x) - g(y)| < \varepsilon
$$
Now if $\|f - f_0\| < \min\{1,\delta\}$, $|f(x) - f_0(x)| < \delta$ for all $x$, so
$$
|F(f_0)(x) - F(f)(x)| = |g(f_0(x))-g(f(x))|  < \varepsilon \quad (\forall x \in [0,1])
$$
and therefore $\|F(f) - F(f_0)\| \leq \varepsilon$. Here I use that $f(x),f_0(x) \in B_{1 + \|f_0\|}(0)$ for any $x$.