Let $X$ be a complete and separable metric space, let $G(X) \subset C([0,1],X)$ be the space of continuous curves from $[0,1]$ to $X$ with constant speed, i.e. $$ d(f(t),f(s)) = |t-s| d(f(0), f(1)). $$ Let $S : X^2 \rightarrow 2^{G(X)}$ be such that at every couple of points $(x,y)$ it associates a curve starting at $x$, i.e. $f(0) = x$, and ending at $y$, i.e. $f(1) = y$. We define the preimage of a set $T \subseteq G(X)$ through $S$ as $$ S^{-1}(T) := \lbrace (x,y) \in X^2 : F(x,y) \cap T \neq \emptyset \rbrace. $$

It can be proven that $G(X)$ endowed with the distance $$\delta(f,g) = \sup_{x \in [0,1]} d(f(x), g(x))$$ is still a complete and separable metric space.

Now take $A \subset G(X)$ an open set. Is $S^{-1} (\lbrace A \rbrace) $ an open set? Is it at least a Borel set?

Take now, in the same situation as before, $S : X^2 \rightarrow 2^{G(X)}$ s.t. at every couple of points $(x,y)$ it associates **all the curves** with constant speed starting at $x$, i.e. $f(0) = x$, and ending at $y$, i.e. $f(1) = y$.

In this situation is the preimage through $S$ of an open subset of $G(X)$ open or Borel in $X^2$?

setof all geodesics connecting $x$ and $y$, just read what is written in your post. How are you planning to define continuity if you have no topology on the target space? $\endgroup$4more comments