Christopher wrote:

> There are two ways to turn a preorder into a equivalence relation. 1) the symmetric closure:

> \$a \sim b := a \le b \text{ or } b \le a \$

Alas, that doesn't always work.

**Puzzle.** Find a preorder such that \$$\sim\$$, defined as above, is not an equivalence relation.

> and 2) the symmetric ??:

> \$a \sim b := a \le b \text{ and } b \le a \$

This does give an equivalence relation, and it's very important. We usually write \$$a \cong b\$$ in this case, and say \$$a\$$ and \$$b\$$ are **isomorphic**.

> I think I've read "core" used for this in the context of turning groups commutative.

I haven't heard of this, and I don't see how exactly this is related to maing groups commutative (aka "abelian"). There's a forgetful functor from abelian groups to groups, and it has a left adjoint called ["abelianization"](http://mathworld.wolfram.com/Abelianization.html). There's also another way to take a group and get an abelian group, called taking the ["center"](https://en.wikipedia.org/wiki/Center_(group_theory)). However, this is not a functor!

Anyway, I guess you're talking about the forgetful functor from the category of "sets with equivalence relation" to the category of preorders, and possible left or right adjoints to this. Method 2) of turning preorders into equivalence relations feels like a right adjoint, though I haven't checked.

Here's another way to turn a preorder into an equivalence relation. Start with a preorder. Take its symmetric closure as you did, defining \$$a \sim b\$$ if \$$a \le b\$$ or \$$b \le a\$$. Then take the [transitive closure](https://en.wikipedia.org/wiki/Transitive_closure) of that relation\$$\sim\$$, getting a relation that's both symmetric and transitive.

I believe this method gives the left adjoint to the forgetful functor from the category of "sets with equivalence relation" to the category of preorders... though again, I haven't carefully checked this so don't sue me if I'm wrong!