Is the natural functor Set$\to$Rel conservative? A conservative functor is a functor such that for any morphism $f$ in $C$, $F(f)$ being an isomorphism implies that f is an isomorphism. Rel stands for the category of relations.
My question is, is the natural functor Set$\to$Rel conservative?
To begin with, let's try to "prove" it. Let $F$ be the natural functor sending a set to the same set, and a map to its graph(which is a relation). The identity map corresponds to the diagonal.
If $F(f)\in \text{Hom}_{\textbf{Rel}}(X,Y)$ is an isomorphism, then there exists a $g'\in \text{Hom}_{\textbf{Rel}}(Y,X)$ such that the compositions of them are diagonals. I have trouble in producing a $g\in \text{Hom}_{\textbf{Set}}(Y,X)$ whose composition with $f$ are identities. Maybe it is false.

It is indeed conservative.  In fact, any isomorphism in $\mathbf{Rel}$ is in fact a function and that function is an isomorphism.
To see this assume that $R \subseteq X \times Y$ is an isomorphism in $\mathbf{Rel}$ with inverse $S \subseteq Y \times X$.  Then looking at the composition $RS = \mathrm{id}_X$ we see that the domain of $R$ is $X$.  From $SR = \mathrm{id}_Y$ we get that the range is $Y$.  Similarly the domain and range of $S$ is $Y$ and $X$ respectively.
Now assume $x \in X$ and $y, z \in Y$ such that $xRy$ and $xRz$.  The element $x$ is in the range of $S$ so choose $u \in Y$ such that $uSx$.  Then by definition we have $uSRy$ and $uSRz$ so $SR = \mathrm{id}_Y$ implies $y = u = z$.  Thus $R$ is a function.  By symmetry $S$ is a function and $RS$ is just composition of functions, so $R$ and $S$ are inverses in $\mathbf{Set}$.
