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Stream: learning: questions

Topic: Reflective subcategories stable under pullback

Bernd Losert (Feb 13 2024 at 22:18):

Let F : C → D be a functor and let B be a reflective subcategory of D with inclusion I : B → D. Under what conditions on F will the pullback of F and I be a reflective subcategory of C?

Morgan Rogers (he/him) (Feb 14 2024 at 09:45):

I assume you're building the pseudo-pullback, whose objects are pairs $(b,c)$ such that $Ib \cong Fc$ etc. Certainly the pullback will always give a full subcategory of C, since fully faithful functors are stable under pullback. To attempt to construct a reflection, let $J : D \to B$ be the adjoint to $I$, and consider $c \mapsto (JF(c),c)$. By construction, $IJF(c) \cong F(c)$, so this is a valid object of the pullback, and you can check that this is functorial. I'll also let you check whether this is a valid adjoint ;) I didn't need any conditions on $F$ to define that, so if it works the answer to your question would be "no conditions"!

Bernd Losert (Feb 14 2024 at 10:07):

Nice. Thanks. I guess the trick is to work with pseudopullbacks.

Morgan Rogers (he/him) (Feb 14 2024 at 10:15):

There are plenty of people here who would tell you that's the kind of pullback you "should" be working with for categories, rather than the strict version. Given your other questions, though, the strict pullback will be straightforwardly equivalent to the pseudo-pullback in cases you are interested in, so you'll be able to get a reflector for the strict pullback if you want it.

Bernd Losert (Feb 14 2024 at 10:24):

In my case, C is already a pullback, so doing the pullback again gives you something that is only isomorphic to a reflective subcategory of C.

Morgan Rogers (he/him) (Feb 14 2024 at 10:26):

Sounds pretty safe to me!

Bernd Losert (Feb 14 2024 at 22:10):

Could it be that you are using "subcategory" in the non-strict sense, i.e. the injection is not necessarily an identity-on-objects functor?

Kevin Arlin (Feb 14 2024 at 22:39):

Almost certainly. It's only in particular situations that category theorists even worry about the difference between equivalent categories; virtually never does one worry about the difference between isomorphic categories. You're doing pullbacks here, and the value of a pullback is only even well-defined up to isomorphism, after all!

Rémy Tuyéras (Feb 15 2024 at 11:03):

Morgan Rogers (he/him) said:

I assume you're building the pseudo-pullback, whose objects are pairs $(b,c)$ such that $Ib \cong Fc$ etc. Certainly the pullback will always give a full subcategory of C, since fully faithful functors are stable under pullback. To attempt to construct a reflection, let $J : D \to B$ be the adjoint to $I$, and consider $c \mapsto (JF(c),c)$. By construction, $IJF(c) \cong F(c)$, so this is a valid object of the pullback, and you can check that this is functorial. I'll also let you check whether this is a valid adjoint ;) I didn't need any conditions on $F$ to define that, so if it works the answer to your question would be "no conditions"!

I quickly skimmed through this, so I might miss something here, but I do not see why, given any object in $c \in C$, the relation $IJF(c) \cong F(c)$ holds. It seems that the weak pullback is constructed with very specific objects $c$, namely those with an image $F(c)$ that lands in B and usually, it is $JI$ that allows us to go back where we started.

In general, I see reflective subcategories like open spaces, so intuitively, I would say that we might need some conditions on F, like the inverse image of B by F is still a category. Then, that inverse image is a subcategory of C, and ideally, it is a reflective subcategory (with maybe some more conditions?)

Morgan Rogers (he/him) (Feb 15 2024 at 11:53):

You're right! Whoops. In fact, the construction I described only works if $F$ factors through $I$, which is a silly condition. Nonetheless, the fact that a pullback of a fully faithful functor is fully faithful is true, so it is definitely a subcategory of $C$, the problem is the existence of a reflector.

A reflector for the projection $C \times_D B \to C$ should send $c$ to some object $(l(c),r(c))$ where $l(c) \in C$, $r(c) \in B$, $Fl(c) \cong Ir(c)$ and $C \times_D B((l(c),r(c)),(c',b')) \cong C(c,c')$. I'll unpack that after lunch ;)

Bernd Losert (Feb 15 2024 at 12:06):

I was actually working out the proof last night and I got stuck. Thanks for pointing out the issue @Rémy Tuyéras.

Bryce Clarke (Feb 15 2024 at 12:08):

This doesn't answer the question, but it may be worth mentioning that if $F \colon A \rightarrow B$ is a functor, and $B$ is a reflective subcategory of $C$ with reflector $L \colon C \rightarrow B$, then $A$ is always a reflective subcategory of the pullback $A \times_{B} C$ of $F$ along $L$.

Morgan Rogers (he/him) (Feb 15 2024 at 13:32):

@Bryce Clarke that's more or less what my original erroneous argument proves haha

Morgan Rogers (he/him) (Feb 15 2024 at 13:59):

To finish the thought from earlier, we can apply $J$ to the isomorphism condition to find $JFl(c) \cong JIr(c) \cong r(c)$, so we might as well take that as the definition of $r(c)$. We can actually eliminate the second argument altogether thanks to full faithfulness of $I$, so $C \times_D B((l(c),JFl(c)),(c',b')) \cong C(l(c),c')$. That's not news, since I already pointed out that the pullback was (equivalent to) a subcategory of $C$! So in the end I don't have any neat conditions for reflectiveness of the subcategory from first principles...

Bernd Losert (Feb 16 2024 at 07:22):

Just want to mention that if C is already a pullback involving D with F : C → D being one of the projections, then it all works out. I originally asked my question because I was hoping for a more general result or maybe a reference with some theorems.

Bernd Losert (Feb 19 2024 at 18:34):

Bernd Losert said:

Just want to mention that if C is already a pullback involving D with F : C → D being one of the projections, then it all works out. I originally asked my question because I was hoping for a more general result or maybe a reference with some theorems.

I take this back. Actually, it does not work out they way I thought. Dang...

Rémy Tuyéras (Feb 19 2024 at 18:39):

@Bernd Losert Even though I feel like you want something general, my intuition would be to assume at least a set of generators in your categories. The reason for this is that you want to construct a reflector, and in practice, this kind of functors is best described as a colimit of generators

Bernd Losert (Feb 19 2024 at 18:41):

Found something akin to what I want: https://ncatlab.org/nlab/show/reflective+sub-%28infinity%2C1%29-category#cocartesian_fiber_over_a_reflective_subcategory

Bernd Losert (Feb 19 2024 at 18:43):

Now, if I remember correctly, a cocartesian fiber is a special case of cotopological functor.