What is the most natural way to make a relation into an equivalence relation? Let $S$ be a set and $R \subseteq S\times S$ be a binary relation on $S$ which is reflexive and symmetric.
Method #1: Say $x\sim y$ iff there exists some $s_0,\ldots, s_n\in S$ such that $s_0 = x$, and $s_n = y$, and $s_iRs_{i+1}$ for all $i$.
Method #2: Say $x\sim y$ iff $xRy$ and for all $z\in S$ such that $yRz$ we have $xRz$.
Both methods produce a, potentially different, equivalence relation. Let $\textsf{RSRel}$ be the category whose objects are sets equipped with a reflexive and symmetric relation, and morphisms are relation-preserving functions. Similarly one can define $\textsf{Equiv}$ to be the category of sets equipped with equivalence relations. Clearly there is a forgetful functor $\textsf{Equiv} \to \textsf{RSRel}$.

Q: Does this functor have an adjoint? If so, is it given by any of the methods above?


Q: What happens when $R$ is "just" a relation, i.e. not necessarily reflexive nor transitive? What is the adjoint to the forgetful functor $\textsf{Equiv} \to \textsf{Rel}$?

Edit to add context: I thought about this question when I was trying to work out what the pushout in the category of sets should be (Exercise I.5.12. in Algebra: Chapter 0). I knew I had to quotient out by this reflexive and symmetric relation but I couldn't because it was not transitive! Eventually I came up with Method #1 which did the trick. I don't know a whole lot about adjoints (not enough to answer my own questions) but I'm curious about whether these (or other) methods are "natural" in any way.
 A: Relations on a set $X$ are just subsets of $X \times X$ and are ordered by inclusion, so it's simpler to work inside the poset of all such subsets for a given $X$. Given any relation $R$ we can consider the intersection of all equivalence relations containing it; intersections of equivalence relations are equivalence relations, so this is an equivalence relation, and it's the smallest equivalence relation containing $R$ which makes it the left adjoint of the inclusion of equivalence relations (on $X$) into all relations (on $X$). It's the equivalence relation generated by $R$.
It can be described more explicitly as follows: thinking of $R$ as describing a directed graph with vertex set $X$, we first convert each directed edge into an ordinary edge (which forces symmetry), then construct the equivalence relation given by sitting in the same path component (which forces reflexivity and transitivity). Alternatively we can phrase this in terms of zigzags.
This construction is the one relevant to constructing pushouts; to quotient by an arbitrary relation you quotient by the equivalence relation it generates.
The question now arises whether the inclusion of equivalence relations into all relations has a right adjoint. But it can't, because the union of equivalence relations is not an equivalence relation in general (it's not necessarily transitive), which means that the inclusion doesn't preserve colimits (joins). So I'm not sure how to generalize your second construction, but it's not the one relevant to constructing pushouts.
