Let $X$ and $Y$ be separable complete metric spaces (if necessary, they may be assumed to be compact). Let $R\subset X\times Y$ be a closed subset such that the projection of $R$ to $X$ is onto. 


**Is it true that there exists a Borel measurable map $f\colon X\to Y$ such that $(x,f(x))\in R$ for every $x\in X$?**


If this is not true in general, are there known sufficient conditions for its existence?



Existence of a measurable map between metric spaces
For a compact space $Y$ the answer is affirmative, but in general case of Polish space $Y$ it is negative. 


Results yielding nice selections of relations $R$ are known in Descriptive Set Theory as Uniformization Theorems, see Section 18 of the standard textbook [A.Kechris, Classical Descriptive Set Theory, Springer, 1995]. 


According to Theorem 18.18 of Arsenin, Kunugui (in this section), for Polish spaces $X,Y$ and a Borel set $R\subset X\times Y$ with all sections $R\_x=\{y\in Y:(x,y)\in R\}$, $x\in X$, $\sigma$-compact and non-empty, there exists a Borel function $f:X\to Y$ whose graph is contained in $R$. 


On the other hand, by Exersice 18.9 in the book of Kechris, for the Polish spaces $X=Y=\omega^\omega$ there exists a closed subset $R\subset X\times Y$ such that for every $x\in Y$ the section $R\_x$ is uncountable, but $R$ admits no Borel uniformization (i.e., no Borel function $f:X\to Y$ whose graph is contained in $R$). Yet by Jankov, von Neumann Uniformization Theorem, $R$ always has a $\sigma(\Sigma\_1^1)$-measurable uniformization.


