This question seems very basic but I have no clue how to show this statement nor have I been able to find some references for it.
Let $X$ and $Y$ be two uniform Hausdorff spaces (i.e. completely regular topological Hausdorff spaces). Consider a finite collection of pairs $(x_i, y_i)$, $i=1,\dots,N$ where $x_i \in X$ and $y_i \in Y$ with $x_i \neq x_j$ whenever $i \neq j$.
Does there always exist a continuous (or even uniformly continuous) function $f: X \to Y$ with $f(x_i) = y_i$ for all $i=1,\dots,N$?
When $Y$ is a (real) vector space it seems straight forward to construct such a function by interpolation. But how to proceed in the general case of uniform/topological spaces?
I would be grateful for any hint how to proof such a statement as well as any pointers to relevant literature. On the other hand, if there is a counterexample, under which assumptions does the statement hold?
Find pairwise disjoint $U_i\ni x_i$ and $f_i\colon X\to [0,1]$ with $f_i(x_i)=1$ and $=0$ outside $U_i$. We obtain a map to a star graph space $Z$ consisting of $n$ copies of $[0,1]$ glued together at their $0$-ends, namely by mapping $x\in X$ to the point $f_i(x)$ on the $i$th copy of $[0,1]$ if $x\in U_i$, and map to the star centre otherwise. This is possible for all allowed $X$, so the actual question is: For what $Y$ can we find a continuous map $Z\to Y$ such that the ends are mapped to given points $y_i$?
Necessarily, the image of $Z$ is path connected. On the other hand, if $Y$ is path connected, then we can pick $z\in Y$ and paths from $z$ to the $y_i$ and combine these to obtain the desired $Z\to Y$.