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Stream: theory: category theory

Topic: Truncation functor


view this post on Zulip Jens Hemelaer (Nov 10 2020 at 14:06):

@Reid Barton gave some interesting point of view here on how to recover results in 1-category theory from the corresponding results in (,1)(\infty,1)-category theory.

I understand that there is some truncation functor τ:SSets\tau : \mathcal{S} \to \mathbf{Sets}, where S\mathcal{S} is the (,1)(\infty,1)-category of spaces and Sets\mathbf{Sets} is the (,1)(\infty,1)-category of sets.

Does this truncation functor preserve colimits and limits? I see here on nlab that it is left adjoint to the inclusion functor the other way?

Similarly, there should be a truncation functor τ:(,1)CatCat\tau : (\infty,1)-\mathcal{Cat} \to \mathbf{Cat}, from the (,2)(\infty,2)-category (?) of (,1)(\infty,1)-categories to the (,2)(\infty,2)-category of 1-categories. Does this functor preserve the same kind of colimits and limits?

If C\mathcal{C} and D\mathcal{D} are (,1)(\infty,1)-categories, and Fun(C,D)\mathbf{Fun}(\mathcal{C},\mathcal{D}) is the (,1)(\infty,1)-category of functors between them, then is there an equivalence of categories τ(Fun(C,D))Fun(τ(C),τ(D))\tau(\mathbf{Fun}(\mathcal{C},\mathcal{D})) \simeq \mathbf{Fun}(\tau(\mathcal{C}),\tau(\mathcal{D}))?

view this post on Zulip Rune Haugseng (Nov 10 2020 at 17:46):

Both truncations preserve colimits (since they are left adjoints) and also products, but not limits in general. The truncation of infinity-categories definitely doesn't preserve functor categories though. In fact I believe you can recover an infinity-category C from the homotopy categories (truncations) of Fun(I, C) for all small ordinary categories I - this is more or less the (pre)derivator approach to higher categories.

view this post on Zulip Jens Hemelaer (Nov 10 2020 at 20:40):

Thank you! I'll have a look at the (pre)derivator approach.

view this post on Zulip Reid Barton (Nov 10 2020 at 20:45):

To recover 1-categorical results from (,1)(\infty,1)-categorical ones you should apply the other adjoint, the inclusion functor of sets in spaces. This commutes with limits and exponentials and so, for example, the (,1)(\infty, 1)-category of functors between two 1-categories is the same as the 1-category of functors between them.

view this post on Zulip Reid Barton (Nov 10 2020 at 20:47):

Likewise, the condition of a diagram being a (co)limit in a 1-category is the same as it being a (co)limit when it is viewed as an (,1)(\infty,1)-category, again because limits in sets agree with limits in spaces.

view this post on Zulip Reid Barton (Nov 10 2020 at 20:50):

If you have a way of building new (,1)(\infty,1)-categories from old ones which involves colimits then it might take you outside the image of the 1-categories, but that isn't common.

view this post on Zulip John Baez (Nov 10 2020 at 20:57):

The homotopy colimit of a set acted on by a group is a groupoid; this shows up all over the place in physics and geometry, and this is why @Joachim Kock said:

You can kill symmetries by dividing out by them, but their ghosts will haunt you forever.

view this post on Zulip John Baez (Nov 10 2020 at 21:06):

And this phenomenon keeps pushing us up the dimensional ladder.

view this post on Zulip Jens Hemelaer (Nov 11 2020 at 08:57):

John Baez said:

The homotopy colimit of a set acted on by a group is a groupoid; this shows up all over the place in physics and geometry, and this is why Joachim Kock said:

You can kill symmetries by dividing out by them, but their ghosts will haunt you forever.

I would compute the homotopy colimit of a diagram F:GSetsF : G \to \mathbf{Sets} as follows. First take the category of elements GF\int_G F of this diagram. This has again a GG-action, and the homotopy colimit is then the 1-categorical colimit of this category under the action of GG.

But I have no idea if/why this gives the correct answer in general.

view this post on Zulip Jens Hemelaer (Nov 11 2020 at 09:07):

Suppose P(C)\mathcal{P}(\mathcal{C}) is the category of presheaves on C\mathcal{C} with values in spaces, for C\mathcal{C} an ordinary category. Then is τ(P(C))PSh(C)\tau(\mathcal{P}(\mathcal{C})) \simeq \mathbf{PSh}(\mathcal{C}), where the latter is the ordinary category of presheaves?

I would prove this by saying that P(C)\mathcal{P}(\mathcal{C}) is the (,1)(\infty,1)-category freely generated under (,1)(\infty,1)-colimits, while PSh(C)\mathbf{PSh}(\mathcal{C}) has the similar 1-categorical universal property.

view this post on Zulip Jens Hemelaer (Nov 11 2020 at 09:22):

Jens Hemelaer said:

Suppose P(C)\mathcal{P}(\mathcal{C}) is the category of presheaves on C\mathcal{C} with values in spaces, for C\mathcal{C} an ordinary category. Then is τ(P(C))PSh(C)\tau(\mathcal{P}(\mathcal{C})) \simeq \mathbf{PSh}(\mathcal{C}), where the latter is the ordinary category of presheaves?

Ok this doesn't make much sense... τ(P(C))\tau(\mathcal{P}(\mathcal{C})) should have the same objects as P(C)\mathcal{P}(\mathcal{C}).

view this post on Zulip Jens Hemelaer (Nov 11 2020 at 11:04):

There is some relevant terminology in the current masterclass by Scholze and Clausen: P(C)\mathcal{P}(\mathcal{C}) is the "animation" of PSh(C)\mathbf{PSh}(\mathcal{C}). So there should be a functor π0:P(C)PSh(C)\pi_0 : \mathcal{P}(\mathcal{C}) \to \mathbf{PSh}(\mathcal{C}) which is left adjoint to the inclusion.

view this post on Zulip Reid Barton (Nov 11 2020 at 12:01):

A note on truncation. It's tempting to think that, when 1-categories are regarded as (,1)(\infty,1)-categories, they are exactly the 0-truncated (,1)(\infty,1)-categories. This is not correct because categories already peek a bit into the higher world. For CC to be a 0-truncated (,1)(\infty,1)-category means that the space of maps from any (,1)(\infty,1)-category to CC is discrete/a set. In particular, this includes the space of maps from the terminal category to CC, which can be identified with the "space of objects" of CC. If CC came from a 1-category then this space can be identified with the maximal groupoid contained in CC, and it is only a set when there are no nontrivial automorphisms of objects in CC.

view this post on Zulip Reid Barton (Nov 11 2020 at 12:05):

That said, the inclusion of 1-categories in (,1)(\infty,1)-categories still has a left adjoint, called the "homotopy category" functor or τ1\tau_1. Informally, its effect is to apply π0\pi_0 to each mapping space. (You could think of it as the change of enrichment along the truncation π0\pi_0 from spaces to sets.)

view this post on Zulip Reid Barton (Nov 11 2020 at 12:06):

By definition, the universal property of the homotopy category is that it has the same functors into any 1-category.

view this post on Zulip Reid Barton (Nov 11 2020 at 12:07):

As you noted, if you apply this to the category P(C)\mathcal{P}(\mathcal{C}) of presheaves of spaces on an ordinary category C\mathcal{C}, you won't get presheaves of sets on C\mathcal{C}. In fact, it's wrong even for C\mathcal{C} the terminal category. Then P(C)\mathcal{P}(\mathcal{C}) is spaces, and its homotopy category is the classical homotopy category of spaces (e.g., the homotopy category of CW complexes).

view this post on Zulip Reid Barton (Nov 11 2020 at 12:10):

What you could do instead in this situation is the following. The (,1)(\infty,1)-category P(C)\mathcal{P}(\mathcal{C}) is (locally) presentable, so it would be better to consider not arbitrary functors out of it but only colimit-preserving ones to cocomplete categories (i.e., left adjoints). Then, we can ask for a universal 1-category with a colimit-preserving functor from P(C)\mathcal{P}(\mathcal{C}). This will be the category of presheaves of sets on C\mathcal{C}.

view this post on Zulip Reid Barton (Nov 11 2020 at 12:14):

In general, this 1-category can be computed as the tensor product with the presentable (,1)(\infty,1)-category Set\mathbf{Set}. Set\mathbf{Set} can be built as a presentable (,1)(\infty,1)-category by imposing the relation ⨿⨿* \amalg_{* \amalg *} * \xrightarrow{\sim} *, so tensoring with Set\mathbf{Set} imposes the relation A⨿A⨿AAAA \amalg_{A \amalg A} A \xrightarrow{\sim} A for every AA. Informally this means that "giving one path in the space of maps from AA to XX is the same as giving two paths", i.e., the space of maps from AA to XX is actually a set, i.e., we obtain a 1-category.

view this post on Zulip Reid Barton (Nov 11 2020 at 12:31):

In the case of P(C)\mathcal{P}(\mathcal{C}) the claim can be verified directly: giving a colimit-preserving functor from P(C)\mathcal{P}(\mathcal{C}) to a 1-category D\mathcal{D} is equivalent to giving an ordinary functor from C\mathcal{C} to D\mathcal{D}, which is in turn equivalent to giving a colimit-preserving functor from the category of presheaves of sets on C\mathcal{C} to D\mathcal{D}.

view this post on Zulip Jens Hemelaer (Nov 11 2020 at 13:04):

Thanks a lot!

So can you recover the 1-Yoneda Lemma from the (,1)(\infty,1)-Yoneda Lemma by using the diagram below?

diagram.png

If F\mathcal{F} is a sheaf with values in sets, then the composition π0Hom(,F)\pi_0 \circ \mathrm{Hom}(-,\mathcal{F}) preserves arbitrary limits, so it factors uniquely through PSh(C)\mathbf{PSh}(\mathcal{C}). The triangle on the left commutes by the (,1)(\infty,1)-Yoneda Lemma, and then the commutation of the outer diagram gives the ordinary Yoneda Lemma.

view this post on Zulip Reid Barton (Nov 12 2020 at 15:22):

This sounds basically correct but maybe it's better to put Sets\mathbf{Sets} where Grpd\infty-\mathbf{Grpd} is, and write the top π0\pi_0 as (π0)op(\pi_0)^\mathrm{op}. You could add a functor from Sets\mathbf{Sets} down to Grpd\infty-\mathbf{Grpd}. Then all the functors not involving Cop\mathcal{C}^\mathrm{op} are right adjoints.

view this post on Zulip Reid Barton (Nov 12 2020 at 15:24):

The point is that Hom(,F)\mathrm{Hom}(-, \mathcal{F}) factors through the inclusion of sets in spaces and is still limit-preserving, and both of those facts are because limits in sets are the same as limits in spaces (and homs into F\mathcal{F} can be computed as a limit of the values of F\mathcal{F}).

view this post on Zulip Jens Hemelaer (Nov 12 2020 at 18:13):

Thanks!