Let $X\_1, X\_2$ be non-empty sets and let $R\subseteq X\_1\times X\_2$ such that for all $x\in X\_1$ there is $y\in X\_2$ such that $(x,y)\in R$.


Are there topologies $\tau\_i$ on $X\_i$ for $i=1,2$ and a continuous function $f:X\_1\to X\_2$ such that $R$ is the topological closure of $\text{graph}(f)$, where $\text{graph}(f) = \bigl\{\bigl(x,f(x)\bigr) \, : \, x\in X\_1\bigr\}$?



Closure of the graph of a function
Take $X\_1=X\_2=\{1,2,3\}$ and $R=\{(x,y):x\neq y\}$. Then the diagonal is open as a complement of the closed subset $R$. As an open subset, the diagonal is a union of products $A\_i\times B\_i$ of some open subsets $A\_i\subseteq X\_1$ and $B\_i\subseteq X\_2$. But these must be singletons so both $X\_i$'s are discrete. Therefore any graph of a function is closed - a contradiction since $R$ is not a graph.


