Let $r$ be a binary relation. Can we assert that there are (monovalued) functions $f$ and $g$ such that
- $r = f\circ g^{-1}$?
- $r = f^{-1}\circ g$?
Here $\circ$ is the relational composition, that is $$q\circ p = \{ (x,y) \mid \exists t:((x,t)\in p\land(t,y)\in q) \}$$ for every binary relations $p$, $q$.
You can always write them in the first way : if $r\subset X\times Y$, take $g:r\to X$ to be the first projection and $f:r\to Y$ to be the second. Then $g^{-1}=(x,(x,y))\in X\times r$, so $$f\circ g^{-1}=\{(x,y)\mid \exists (x',y')\in r : (x,(x',y'))\in g^{-1}\wedge ((x',y'),y)\in f\}=r,$$ because the first condition is equivalent to $x=x'$ and the second to $y=y'$.
On the other hand, the second way does not always work. Indeed, if $f,g$ are two functions $X\to Y$, then $$f^{-1}\circ g=\{(x,x')\mid \exists y :(x,y)\in g \wedge (y,x')\in f^{-1}\}=\{(x,x')\mid g(x)=f(x')\}.$$ In particular, if we can find $a,b\in X$ such that $(a,a),(a,b),(b,b)$ are all in $r$, then we must have $$f(a)=g(a)=f(b)=g(b),$$ so that we must also have $(b,a)\in r$. In particular, any relation that is reflexive but not symmetric, such as a non-trivial partial order, cannot be of this form.