Can we define a reverse mapping for the $\phi : Y \rightarrow C(X, Y)$, as $\phi(y) = y' $, where $y' $ is a constant map in $C(X, Y)$.

Question

For a continuous map $\phi : Y \rightarrow C(X, Y)$, as $\phi(y) = y' $, where $y' $ is a constant map in $C(X, Y)$. Can we define a reverse mapping as $\phi' : C(X, Y) \rightarrow Y$ for a metric space $Y$. If yes how can it be defined? And suppose $C(X, Y)$ is endowed with uniform topology.

Answer 1

Ideas: if $X$ is a connected space and $Y$ is totally disconnected (like $Y=\Bbb Q$ or $Y=C$ the Cantor set etc.) then $C(X,Y) = \phi[Y]$, i.e. all continuous maps from $X$ to $Y$ are constant, and $\phi$ has an obious inverse that is also continuous (in most natural topologies that we could put on $C(X,Y)$).

Otherwise I don't really see a natural map $C(X,Y) \to Y$, unless in specific situations like $X$ pseudocompact and $Y$ an ordered space, where $f \to \max f[X]$ or $f \to \min f[X]$ seems natural. I think it will be continuous in the uniform topology, though I haven't checked the details.

Even more specific, if $Y=\Bbb R^n$ and $C(X,Y)$ has such a topology that the set of constant maps $\phi[Y]$ is closed in it and $C(X,Y)$ is normal, then Tietze applies to show there is a continuous extension of $\phi^{-1}: \phi[Y] \to Y$ to $C(X,Y)$.