Adjoint to a functor $\textbf{PoSets}\rightarrow\textbf{PreOrd}$

Question

i have a question about adjoints in category Theory. Let $\textbf{Posets}$ the category of Posets (thus Sets with binary relation $\leq$ which is reflexiv, transitiv and antisymmetric) and let $\textbf{PreOrd}$ the category of preoderer sets (thus sets with a binary relation $\leq$ which is reflexiv and transitiv).

Thus we can make an obvious forgetful functor $G:\textbf{Posets}\rightarrow\textbf{PreOrd}$ which "forget" the antisymmetry of the poset. The exercise is now the following:

Show that the functor G has a left adjoint.

My idea was the following. We have to do this in some steps. First we have to make a functor $F:$ PreOrd $\rightarrow$ Posets. Is it correct to define $F(X,\leq)=(X,\leq)$ by adding the property of antisymmetry, or can this not be done?

In the next step we have to prove the bijective correspondence between $F(X)\rightarrow Y$ and $X\rightarrow G(Y)$, thus given $f:F(X)\rightarrow Y$ construct $\bar{f}:X\rightarrow G(Y)$ and also for a given $g:X\rightarrow G(Y)$ an arrow $\bar{g}:F(X)\rightarrow Y$ such that $\bar{\bar{f}}=f$ and $\bar{\bar{g}}=g$.

But how to make a good construction? Can someone help me with this exercise?

Thanks a lot.

Answer 1

You have to make precise "by adding the property of antisymmetry". Define $F(X,\leq) = (X/\sim,\leq)$ where $x \sim y \Leftrightarrow x \leq y \leq x$ and $[x] \leq [y]$ iff $x \leq y$. Show that $(X,\leq) \to G(X/\sim,\leq)$ is universal.

Answer 2

Hint:

Let $\left(X,\leq\right)$ be a preorder and define the relation $\sim$ on $X$ by $x\sim x'\iff x\leq x'\wedge x'\leq x$. It is an equivalence-relation. Define $\preceq$ on $X/\sim$ by $\left[x\right]\preceq\left[x'\right]\iff x\leq x'$. Then $\preceq$ is well-defined and $\left(X/\sim,\preceq\right)$ is a poset. Try it with a functor $F:\mathbf{Preord}\rightarrow\mathbf{Posets}$ that sends object $\left(X,\leq\right)$ to object $\left(X/\sim,\preceq\right)$.