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Let $E$ be a Banach space, $X$ a locally-compact metric space, equip $C(X,E)$ with the compact-open topology. Let $F:E \rightarrow E$ and consider the induced map $F_{\star}(f):=F \circ f$ on $C(X,E)$. When is $F_{\star}$ open?

Update More generally, what is an example of an open mapping $G:C(X,E)\to C(X,E)$ which is not a homeomorphism (possibly not continuous)?

Update: I think this should work, but, I feel that something is wrong with the argument...

Let $F:E\rightarrow E$ be open and admit a continuous-left inverse. For every compact $K\subseteq X$ and over open $U\subseteq E$ (both non-empty) we know that $\{f\in C(X,E):f(K)\subseteq U\}$ is a basic open subset of $C(X,E)$ for this topology. Then $$ F_{\star}(\{f\in C(X,E):f(K)\subseteq U\})= \{g \in C(X,E): (\exists f \in C(X,E))\, g=F\circ f \mbox{ and } g(K)\subseteq F(U)\} = \{g \in C(X,E): g(K)\subseteq F(U)\} , $$ but since $F$ is open then $F(U)$ is open. Hence, $F_{\star}(\{f\in C(X,E):f(K)\subseteq U\})$ is a basic open subset of $C(X,E)$ for this topology.

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  • $\begingroup$ Do you mean $\exists g \in C(X, E), f = F \circ g$ ? $\endgroup$ Jul 20, 2020 at 9:58
  • $\begingroup$ Oops, yes indeed (or else this would be a wildly different question, with unclear reference to g). Thank you for pointing that out. $\endgroup$
    – ABIM
    Jul 20, 2020 at 10:00
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    $\begingroup$ For $X = \mathbb R$, the constant function equal to 1 is in your set but its hard to believe that there is a neighborhood around it of functions bounded by $1$. $\endgroup$ Jul 20, 2020 at 10:06
  • $\begingroup$ @InfiniteLooper Right. I focused the question then to only the generalized (second part); i.e.: when is post-composition with $F$ open? $\endgroup$
    – ABIM
    Jul 20, 2020 at 10:09
  • $\begingroup$ Why is every continuous function $g$ with $g(K) \subseteq F(U)$ a composition $g = F \circ f$ with $f$ continuous? In particular, I don't see why we can solve $g = F \circ f$ even on $K$ (with $f$ continuous); but, if that's not an issue, what if $g(K) \subseteq F(U)$ but not $g(X) \subseteq F(X)$? $\endgroup$
    – LSpice
    Jul 20, 2020 at 10:55

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