# Forgetful functor from Pos to Set does not have a right adjoint

I'm trying to show that the forgetful functor from $$Pos$$ to $$Set$$ does not have a right adjoint, by showing that it does not preserve coequalizers.

The hint in the lecture notes I am studying, suggests looking at the coequalizer of the following two maps from $$\mathbb{Q}$$ to $$\mathbb R$$, namely the inclusion and the constant zero map.

I think the coequalizer of these in $$Set$$ will be the quotient Map from $$\mathbb R$$ to $$\mathbb{R/Q}$$, but I can't seem to figure out what the coequalizer will be in $$Pos$$, and why it will not be preserved.

Any hints/solutions will be appreciated.

Thank you.

## 1 Answer

In the category of posets, a coequalizer $$X \rightrightarrows Y$$ is calculated by first taking the set-theoretic coequalizer $$q: Y \to Z$$. Then we make $$Z$$ into a preorder by giving it the smallest preorder such that $$q$$ is order-preserving (this is by the way how we calculate preorders in the category of preorders). Finally we make $$Z$$ into a partial order by dividing out the equivalence relation $$z \sim z' \quad \Longleftrightarrow \quad z \leq z' \text{ and } z' \leq z.$$ So let's do this for the inclusion $$i: \mathbb{Q} \hookrightarrow \mathbb{R}$$ and the zero map $$0: \mathbb{Q} \to \mathbb{R}$$. The set-theoretic coequalizer is indeed $$Z = \mathbb{R}/\mathbb{Q}$$. Then the preorder says that $$[x] \leq [y]$$ precisely when $$x \leq y$$ for $$x,y \in \mathbb{R}$$ (and $$[x]$$ denotes the equivalence class of $$x$$). For any $$x, y \in \mathbb{R}$$ (wlog $$x \leq y$$) we can find rationals $$q_1, q_2 \in \mathbb{Q}$$ such that $$q_1 \leq x \leq y \leq q_2$$. So: $$[q_1] \leq [x] \leq [y] \leq [q_2],$$ and since $$[q_1] = [q_2]$$ we have $$[x] = [y]$$. We thus conclude that by making $$Z$$ into a partial order, we are left with only one element. So the coequalizer is $$t: \mathbb{R} \to 1$$, and is indeed different from $$\mathbb{R} \to \mathbb{R} / \mathbb{Q}$$ in $$\mathbf{Set}$$.