• A groupoid can be regarded as a small category in which every arrow is an isomorphism
  • A monoid can be regarded as a small category with only a single object
  • A preorder can be regarded as a small category in which each hom-set has at most one element.

Hence in each case there is an inclusion functor between the corresponding category and the category of small categories. I wonder which of these inclusion functors have a left adjoint, and how to construct them.

To map an arbitrary small category to a groupoid, I would like to just omit all arrows which are not isomorphisms. Even so this seems to be a canonical construction, it doesn't seem to be a functor. After all a functor should map every arrow somewhere, but what to do with the omitted arrows? (Looks like I fell into a trap, the "internal" arrows are not the relevant morphisms here.) But maybe there is a completely different ("Grothendieck group" like) construction, which actually works and is a left adjoint?

To map an arbitrary small category to a monoid, I would like to single out one object. However, this doesn't seem to be a canonical construction. But perhaps the following construction works. If the category has more than a single object, then I add ("adjoin") a new bottom element (or "morphism"), and define every undefined composition to give the bottom element. The resulting binary operation is total and associative, so this seems to be a canonical construction. It maps each object to the single monoid object and each morphism to an element of the monoid, so it seem to be a functor. But is this functor a left adjoint of the inclusion functor?

To map an arbitrary small category to a preorder, I could just replace each non-empty hom-set by a hom-set with a single element. This construction should give the left adjoint functor of the inclusion functor, no?

  • $\begingroup$ Freyd's theorem constructs it explicitly. $\endgroup$ – user40276 Feb 2 '14 at 16:16
  • $\begingroup$ The inclusion functor $\mathbf{Grp}\to\mathbf{Cat}$ doesn't preserve colimits... $\endgroup$ – Oskar Feb 2 '14 at 16:20
  • $\begingroup$ @MartinBrandenburg I mean if it satisfies the conditions (which apparently holds), Freyd's theorem shows an explicitly construction. $\endgroup$ – user40276 Feb 2 '14 at 16:21
  • $\begingroup$ Yes, if you read the proof carefully there is explicit construction. So, apparently, for grupoids it does not hold. $\endgroup$ – user40276 Feb 2 '14 at 16:24
  • $\begingroup$ @Oskar Could please show a counter-example? I'm very interested in it. $\endgroup$ – user40276 Feb 2 '14 at 16:26

You have correctly described the right adjoint of $\mathsf{Gpd} \to \mathsf{Cat}$. It maps a small category to its core. It is a functor $\mathsf{Cat} \to \mathsf{Gpd}$. Given a functor between small categories, it maps isomorphisms to isomorphisms, hence restricts to a functor between the cores.

The functor $\mathsf{Mon} \to \mathsf{Cat}$ has no right adjoint because it does not preserve colimits. In fact, already the initial object is not preserved (the initial monoid is $\{1\}$, whereas the initial category is empty). However, this functor has a left adjoint: We map a category $C$ to the free monoid generated by $\mathrm{Mor}(C)$ modulo the relations $[f] * [g] = [f \circ g]$ if $f,g \in \mathrm{Mor}(C)$ and $\mathrm{cod}(g)=\mathrm{dom}(f)$. In some sense this is the category with all objects contracted to a single object.

The functor $\mathsf{PreOrd} \to \mathsf{Cat}$ has a left adjoint: It maps a small category $C$ to the preorder on $\mathrm{Ob}(C)$ given by $X \leq Y$ iff there is a morphism $X \to Y$. But it has no right adjoint.

  • $\begingroup$ Nice, but what about the left adjoint of $\mathbf{Gpd} \rightarrow \mathbf{Cat}$? $\endgroup$ – user40276 Feb 2 '14 at 17:09
  • 4
    $\begingroup$ It also exists. One adjoins inverses to every morphism. $\endgroup$ – Martin Brandenburg Feb 2 '14 at 17:20

Besides Martin's answer, you might enjoy adding - or should I say adjoining :-) - another adjunction to your already fine collection.

It's actually an adjunction quadruplet!

$$Comp \dashv D \dashv Ob \dashv Ind$$


$Comp: \mathsf{Cat} \to \mathsf{Set}$, $C \mapsto $set of connected components of $C$

$D: \mathsf{Set} \to \mathsf{Cat}$, $S\mapsto $discrete category of $S$

$Ob: \mathsf{Cat} \to \mathsf{Set}$, $C \mapsto $set of objects of $C$

$Ind: \mathsf{Set} \to \mathsf{Cat}$, $S\mapsto $indiscrete category of $S$

$Ind(S)$ is the category with objects $S$ and exactly one arrow in every hom-set. It is a preorder, a groupoid and, if $S$ is not-empty, it is equivalent to the terminal category

All this can be read in CWM 2nd ed. page 90 exercise 9.

  • $\begingroup$ @ThomasKlimpel Good catch! I edited. $\endgroup$ – magma Feb 4 '14 at 13:07
  • $\begingroup$ I would call $TC(S)$ the indiscrete category on $S$. $\endgroup$ – Martin Brandenburg Feb 4 '14 at 13:57
  • $\begingroup$ @MartinBrandenburg Thank you, I edited. $\endgroup$ – magma Feb 4 '14 at 14:36
  • $\begingroup$ These adjunctions are also adjunctions between $\mathsf{PreOrd}$ and $\mathsf{Set}$. All functors factor through the adjoint functors between $\mathsf{Cat}$ and $\mathsf{PreOrd}$. Only $D \dashv Ob$ is a "genuinely new" adjunction between $\mathsf{Set}$ and $\mathsf{Cat}$, because the corresponding functors between $\mathsf{PreOrd}$ and $\mathsf{Cat}$ are not adjoint in the same sense. $\endgroup$ – Thomas Klimpel Feb 5 '14 at 8:21

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