Michael wrote:

> I have a question about the forgetful functor \\(\mathbf{Preord} \stackrel{R_2}{\longrightarrow} \mathbf{Poset} \\). How would this functor forget 2-cycles? Since we are keeping the objects, it seems to me there can be a choice of which arrow to forget but don't think it can be mapped to the other arrow coming back since source and target won't match. Does the whole cycle just get forgotten i.e. mapped into an identity?

I'm having a bit of trouble understanding this, but as others have hinted:

The difference between a poset and a preorder is that a poset is a preorder where isomorphic objects are equal, i.e., one where

$$ x \le y \text{ and } y \le x \implies x = y $$

So, if you're trying to dream up a functor

\[\mathbf{Preord} \stackrel{R_2}{\longrightarrow} \mathbf{Poset} \]

the natural thing to make it do is _force isomorphic objects to be equal_.

To do this, you take your preorder, say \\(X\\), and form a new set \\(R_2(X)\\) whose elements are _isomorphism classes_ of elements of \\(X\\), where \\(x,y\in X\\) are **isomorphic** iff

$$ x \le y \text{ and } y \le x $$

Let \\([x]\\) stand for the isomorphism class of \\(x\in X\\), so that

$$ x \le y \text{ and } y \le x \iff [x] = [y] $$

Let \\(R_2(X)\\) be the set of these isomorphism classes. Then:

* How can we make \\(R_2(X)\\) into a poset using the fact that \\(X\\) was a preorder?

* Can we define \\(R_2\\) on morphisms, too? Given a monotone function between preorders, say \\(f: X \to X'\\), we want to define a monotone function \\(R_2(f) : R_2(X) \to R_2(X')\\).

* Can we check that \\(R_2\\) is a functor? We want to show that it preserves composition and identities.

> I have a question about the forgetful functor \\(\mathbf{Preord} \stackrel{R_2}{\longrightarrow} \mathbf{Poset} \\). How would this functor forget 2-cycles? Since we are keeping the objects, it seems to me there can be a choice of which arrow to forget but don't think it can be mapped to the other arrow coming back since source and target won't match. Does the whole cycle just get forgotten i.e. mapped into an identity?

I'm having a bit of trouble understanding this, but as others have hinted:

The difference between a poset and a preorder is that a poset is a preorder where isomorphic objects are equal, i.e., one where

$$ x \le y \text{ and } y \le x \implies x = y $$

So, if you're trying to dream up a functor

\[\mathbf{Preord} \stackrel{R_2}{\longrightarrow} \mathbf{Poset} \]

the natural thing to make it do is _force isomorphic objects to be equal_.

To do this, you take your preorder, say \\(X\\), and form a new set \\(R_2(X)\\) whose elements are _isomorphism classes_ of elements of \\(X\\), where \\(x,y\in X\\) are **isomorphic** iff

$$ x \le y \text{ and } y \le x $$

Let \\([x]\\) stand for the isomorphism class of \\(x\in X\\), so that

$$ x \le y \text{ and } y \le x \iff [x] = [y] $$

Let \\(R_2(X)\\) be the set of these isomorphism classes. Then:

* How can we make \\(R_2(X)\\) into a poset using the fact that \\(X\\) was a preorder?

* Can we define \\(R_2\\) on morphisms, too? Given a monotone function between preorders, say \\(f: X \to X'\\), we want to define a monotone function \\(R_2(f) : R_2(X) \to R_2(X')\\).

* Can we check that \\(R_2\\) is a functor? We want to show that it preserves composition and identities.