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 1


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 $\mathbf{Set}$.


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