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Ben Sprott (Feb 08 2021 at 18:09):There is a known Free Groupoid Functor from Graphs to the category of Groupoids. It is given as the example in the nlab article on free Groupoids. I would like to create such an adjunction between the category of Groupoids and the category of, perhaps finite, Sets. I just picture doing this by taking any set or set of sets and attaching every isomorphism in sight. Then, you just see the sets and isomorphisms as objects and morphisms of a given Groupoid. Does this work? Could it work if we restrict the category of Groupoids to contain only Groupoids that are categories that are finely presented?
Nathanael Arkor (Feb 08 2021 at 20:23):Do you mean you want to add an isomorphism between every pair of elements of the set, considered as objects of the category?
Nathanael Arkor (Feb 08 2021 at 20:25):The forgetful functor from categories to sets has a right adjoint, given by sending each set to the codiscrete category on that set (which is precisely adding isomorphisms between each pair of elements). This functor factors through the category of groupoids. (Note that the categories in the image of this functor are all equivalent to the terminal category, though.)
Ben Sprott (Feb 08 2021 at 23:19):@Nathanael Arkor
Do you mean you want to add an isomorphism between every pair of elements of the set, considered as objects of the category?
Yes. So the objects of the categories(groupoids) are the sets and the morphisms of the categories are every isomorphism of the set to itself. I am picturing being able to do this with sets of sets and then you have categories with more than one object and you just add to the Category all the endo-isos and all isos between all of the sets.
Ben Sprott (Feb 08 2021 at 23:20):@Nathanael Arkor
The forgetful functor from categories to sets has a right adjoint, given by sending each set to the codiscrete category on that set (which is precisely adding isomorphisms between each pair of elements). This functor factors through the category of groupoids. (Note that the categories in the image of this functor are all equivalent to the terminal category, though.)
So does this mean there is a construction like I'm describing, but it's trivial?
Nathanael Arkor (Feb 08 2021 at 23:21):Ben Sprott said:
Nathanael Arkor
Do you mean you want to add an isomorphism between every pair of elements of the set, considered as objects of the category?
Yes. So the objects of the categories(groupoids) are the sets and the morphisms of the categories are every isomorphism of the set to itself. I am picturing being able to do this with sets of sets and then you have categories with more than one object and you just add to the Category all the endo-isos and all isos between all of the sets.
What you describe here sounds like the category of sets and bijections. But this is a specific category. (Also, "set of sets" doesn't make sense: the collection of all sets forms a proper class, not a set.)
John Baez (Feb 08 2021 at 23:58):While Ben said
Yes.
his answer to Nathanaels' question actually seems to be "No". Remember Nathanael's question:
Do you mean you want to add an isomorphism between every pair of elements of the set, considered as objects of the category?
This one of two famous ways to turn a set into a groupoid. In this way, you take the objects of the groupoid to be elements of your set, and put in one isomorphism from each object to each other object.
John Baez (Feb 09 2021 at 00:00):Since Ben was originally asking about how to get an adjunction between the category of groupoids and the category of finite sets, Nathanael's question was a good one. Nathanael was talking about one of the two famous ways to do this.
John Baez (Feb 09 2021 at 00:04):But then, Ben's comment:
So the objects of the categories (groupoids) are the sets and the morphisms of the categories are every isomorphism of the set to itself. I am picturing being able to do this with sets of sets and then you have categories with more than one object and you just add to the Category all the endo-isos and all isos between all of the sets.
shoots off in a completely different direction. So I can no longer tell what he was asking about.
He seems now to be talking about a specific groupoid: the groupoid with sets as objects and bijections as morphisms.
Ben Sprott (Feb 09 2021 at 20:16):I have been accused many times of going on "fishing expeditions". I get some intuition, usually from my experiences as a physicist, and I try to phrase it mathematically with my tiny toolbox.
@John Baez
He seems now to be talking about a specific groupoid: the groupoid with sets as objects and bijections as morphisms.
Perhaps this better describes what I am thinking of. I may be thinking of the functor from Set to the groupoid of sets and bijections. This functor works by forgetting every function except the isomorphisms. Could this have an adjoint? Does it extend to a functor from Cat to Groupoid which works by forgetting every morphism in every category except the isomorphisms. Does this have an adjoint?
Ben Sprott (Feb 09 2021 at 20:19): Nathanael Arkor (Feb 09 2021 at 20:19):You can't define a functor from by "forgetting noninvertible functions", because a functor has to map every morphism in the domain to a morphism in the codomain. There is, however, a functor in the other direction including sets and bijections into sets and functions. (And this indeed extends to a functor from ).
Nathanael Arkor (Feb 09 2021 at 20:20):And yes, the core gives a right adjoint to this inclusion .
Ben Sprott (Feb 09 2021 at 23:28):Does this adjunction generate a monad on Groupoid? If so, what is that monad? What about on Cat?
Nathanael Arkor (Feb 10 2021 at 00:07):Every adjunction induces a monad on the domain of the left adjoint, and a comonad on the codomain. These functors are very simple, so it would be a good exercise to work out what they are yourself.