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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeJul 22nd 2010

• an expanded Idea-section (check!)

• the evident proposal for the definition of $\left(\infty ,1\right)$-categoical images here (which implies in particular a notion of images of of $n$-functors between $n$-groupoids).

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeJul 22nd 2010

Image of a category appears as a part of a factorization system on Cat. Can this help to get an alternative check on your approach ?

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeJul 22nd 2010

Yeah, somebody should work out what the 2-limit that I am proposing actually boils down to.

But: it’s clearly the right definition. ;-)

• CommentRowNumber4.
• CommentAuthorzskoda
• CommentTimeJul 22nd 2010
• (edited Jul 22nd 2010)

I meant the factorization systems of that kind have refinement into (n+1)-step factorization systems which are categorical analogues of the Postnikov tower, as alluded in the lectures of Baez and Shulman. I find that fascinating…I wish somebody finishes this idea from their lectures with hard theorems.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeJul 22nd 2010
• (edited Jul 22nd 2010)

categorical analogues of the Postnikov tower, as alluded in the lectures of Baez and Shulman. I find that fascinating…I wish somebody finishes this idea from their lectures with hard theorems.

We have hard theorem about the notion of Postnikov tower in an (infinity,1)-category.

The definitions of that, at least, have immediate analogs for $\left(\infty ,n\right)$-categories.

• CommentRowNumber6.
• CommentAuthorzskoda
• CommentTimeJul 22nd 2010
• (edited Jul 22nd 2010)

I know there is much about general Postnikov towers in (infinity,1)-setting. But I was talking about having specifically the statement that there is a (n+2)-step factorization system on the category of weak (n,n)-categories, and that this factorization system is related to some case of Postnikov construction in (infinity,1)-setting. My understanding is that the existence of that (n+2)-step factorization system is known for strict n-categories. In Baez-Shulman paper this is alluded in the treatment of $\left(n,k\right)$-surjectivity. Thus I am looking for a full solution to a very concrete problem alluded in Baez-Shulman exposition. In fact, in written literature, there is no full treatment of the factorization system for strict n-categories, but it is probably just the matter of writing or at worst an undergraduate diploma project.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeJul 22nd 2010
• (edited Jul 22nd 2010)

My understanding is that the existence of that (n+2)-step factorization system is known for strict n-categories. In Baez-Shulman paper this is alluded in the treatment of (n,k)-surjectivity.

My understanding is the the notion of k-surjective functor in their article is the globular reformulation of what simplicially is a morphism that has the right lifting property against the $k$th generating cofibration

$\partial \Delta \left[k\right]\to \Delta \left[k\right]\phantom{\rule{thinmathspace}{0ex}}.$

Namely a functor between strict globular $\infty$-categories is $k$-surjective precisely it if it has the right lifting property against the inclusions of the boundary of the $k$-globe into the $k$-globe

$\partial {G}_{k}\to {G}_{k}$

(See here.)

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeJul 22nd 2010
• (edited Jul 22nd 2010)

And essentially $k$-surjective is something like surjective on the $k$th categorical homotopy groups, where the $k$th-categorical homotopy group is images of ${G}_{k}$ or $\Delta \left[k\right]$ with constant boundary modulo translation along equivalences.

• CommentRowNumber9.
• CommentAuthorzskoda
• CommentTimeJul 22nd 2010

Right. Now the question is if one can make a factorization system out of those as sketched in Baez-Shulman.

• CommentRowNumber10.
• CommentAuthorDavidRoberts
• CommentTimeJul 23rd 2010

I have an idea how to do this. Take for a test case 1-categories. For a category $C$ map $X\to \mathrm{Ob}\left(C\right)$ let $C\left[X\right]$ be the category given by the pullback $\mathrm{codisc}\left(X\right){×}_{\mathrm{codisc}\left(\mathrm{Obj}\left(C\right)\right)}C$ (In other words, objects $X$ and arrows ${X}^{2}{×}_{\mathrm{Obj}\left(C{\right)}^{2}}\mathrm{Mor}\left(C\right)$). Then for a functor $f:D\to C$ there is a factorisation

$D\to C\left[\mathrm{Obj}\left(D\right)\right]\to C\left[\mathrm{im}\left({f}_{0}\right)\right]\to C$

where ${f}_{0}$ is the object-component of $f$.

A similar game can be played with 2-categories, where one can not only define $C\left[X\right]$ for a set $X$, but also a reflexive graph ${X}_{•}=\left({X}_{1}⇉{X}_{0}\right)$ with composition (something like a (pointed magma)-oid :) and a map of such things ${X}_{•}\to {C}_{\le 1}$ for the underlying 1-category ${C}_{\le 1}$ of a 2-category (this works for bicategories too, but let’s stick to strict things). In particular, for the underlying category ${D}_{\le 1}$ of a second 2-category $D$ equipped with a 2-functor $F:D\to C$. In general this is a bicategory, but if ${X}_{•}$ is a category, then $C\left[{X}_{•}\right]$ is a 2-category.

We can then define

$D\to C\left[{D}_{\le 1}\right]\to C\left[{D}_{\le 0}\right]\to C\left[\mathrm{im}\left({F}_{0}\right)\right]\to C$

Each of the things $C\left[?\right]$ expresses that $2\mathrm{Cat}$ is fibred (opfibred?) over various other categories and they can all be defined in terms of pullbacks with variously codiscrete 2-categories(1). This should make it manifest what sort of things are forgotten (stuff, structure etc) at each step. The universal property of the pullbacks helps ensure the uniqueness up to isomorphism of the factorisation.

The general pattern should be clear. There is a functor ${\mathrm{codisc}}_{n+1}:\mathrm{nCat}\to \left(n+1\right)\mathrm{Cat}$ adding to each category a unique arrow between any two parallel n-arrows. For an $n$-functor $D\to C$ define $C\left[{D}_{\le m}\right]={\mathrm{codisc}}_{m+1}\left({D}_{\le m}\right){×}_{\mathrm{codisc}\left({C}_{\le m}\right)}C$ where the $\left(m+1\right)$-codiscrete $\left(m+1\right)$-categories are considered $n$-categories with only identity arrows between dimensions $m+1$ and $n$. Here $m=0,\dots ,n-1$. The successive truncations together with the universal property of the pullback should furnish the functors in the factorisation.

(1) There is a functor ${\mathrm{codisc}}_{2}:{\mathrm{rGrp}}_{\mathrm{comp}}\to \mathrm{Bicat}$ from reflexive graphs with composition to $\mathrm{Bicat}$ by adding to a reflexive graph a unique 2-arrow between any two parallel 1-arrows. Coherence happens automatically. The restriction of ${\mathrm{codisc}}_{2}$ to $\mathrm{Cat}↪{\mathrm{rGrp}}_{\mathrm{comp}}$ lands in $2\mathrm{Cat}↪\mathrm{Bicat}$.

• CommentRowNumber11.
• CommentAuthorDavidRoberts
• CommentTimeJul 23rd 2010

The simplicial version of this should be by using coskeleta. Actually I think the above can be expressed more succinctly by by considering the codiscrete categories of the truncations as steps in the Postnikov tower for $D$ (not really a Postnikov tower except for groupoids, but you know what I mean. It is easy to check these form a tower, where each map in it is taking equivalence classes of the top-dimensional morphisms): and then follow the formal construction of the Moore-Postnikov tower in $\mathrm{Top}$ from the Postnikov tower of one of the spaces involved (can’t remember off the top of my head how this is done, but it’s easy).

• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeJul 23rd 2010
• (edited Jul 23rd 2010)

The page image says that its various definitions are equivalent when they jointly apply. But I think this is false, since the first two (“in terms of subobjects” and “as a left adjoint functor”) deal only with monomorphisms, while the third (“as an equalizer”) always produces a regular monic.

I think the problem is that the notion of “image” really depends on first fixing a notion of subobject, which is different in different contexts. I prefer the first two definitions at image, which can easily be relativized to any notion of “subobject”, and are closely related to factorization systems. On the other hand, the third definition I view as simply a construction of images in the particular case when “subobject” means “regular mono.” I know that some people use “coimage” and “image” to mean specifically quotients of kernels and coquotients of cokernels, but I think I would prefer to view both of those as constructions of particular kinds of images.

Back on the topic of everyone else’s discussion… I don’t quite understand what the question is. Is it whether there is (what I would now prefer to call) a (strong k-epic, k-monic) factorization system on strict n-categories? Or on weak n-categories (what sort)? Or whether such a factorization can be constructed using limits and colimits in some higher-categorical way?

(By the way, I’m currently thinking of using “k-full” for what was called “essentially k-surjective” in Baez-Shulman. The problem with “k-surjective” is that a functor of this sort isn’t literally surjective on k-morphisms, even up to equivalence – only “locally” so in a sense which exactly generalizes the ordinary notion of “full functor.” In particular, the functor out of the empty category is k-full for all k>0. Also, “essentially k-surjective” could be argued to violate the rule that a “1-thing” should be the same as a “thing,” since we often say “essentially surjective” for “essentially surjective on objects”, i.e. 0-full. Thoughts?)

• CommentRowNumber13.
• CommentAuthorDavidRoberts
• CommentTimeJul 23rd 2010

I think Zoran was wondering if the details of the ’layer cake philososphy’ of multi-stage factorisation had been written down for n-functors, in particular, for strict functors and n-cats as a warm-up. One is lead to wonder: what is the definition of an $n$-step factorisation system on a category? One could require uniqueness up to various compatible isomorphisms…

I claim my prescription in #10 would work for weak $n$-categories, I think the only thing to show is that there is a strict pullback of $n$-cats, and that the operation ’identify parallel top-dimensional arrows in an $n$-category’ gives an $\left(n-1\right)$-category and that this is functorial.

I like the idea of k-full. For 1-categories ’2-full’ should mean ’faithful’, for the same reason that a locally full 2-functor between 1-categories is just a faithful 1-functor. But one could modify and say things like (at the cost of syllables!) essentially locally $k$-surjective or some permutation. Or we could throw an ’essentially’ in there, with essentially $k$-full. Or we could go the other way and take $k$-full as the weak notion, and use strictly $k$-full for the actually surjective on parallel $k$-morphisms.

• CommentRowNumber14.
• CommentAuthorMike Shulman
• CommentTimeJul 23rd 2010

I definitely want “k-full” to mean the up-to-equivalence notion by default. Yes, of course 2-full for 1-categories means faithful, and so on.

Is the factorization $D\to C\left[\mathrm{ob}\left(D\right)\right]\to C\left[\mathrm{im}\left({f}_{0}\right)\right]\to C$ supposed to be the 3-step factorization of a functor? It looks to me like the two categories in the middle are equivalent and both give the (eso, ff) factorization.

• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeJul 23rd 2010
• (edited Jul 23rd 2010)

The page image says that its various definitions are equivalent when they jointly apply. But I think this is false,

So what do we do? You are the one with most thoughts on this, so I think it would be good if you edited the entry according to your comments in #12.

• CommentRowNumber16.
• CommentAuthorzskoda
• CommentTimeJul 23rd 2010
• (edited Jul 23rd 2010)

I am not sure about your present notation, but I have checked in all details at some point the 3 step factorization on strict categories and 4 step factorization on strict 2-categories, after reading your paper with John. On the other hand, I do not recall that i cleared out the full axiomatic definition of what n-step factorization systems are, though some wanted properties are immediate. David correctly understood my question (on the other hand he organizes the construction proposal in 10 a bit different than what I remember so I need to rethink carefully; I hope to return to that next week – I have due the Arnold paper for local radio next week).

• CommentRowNumber17.
• CommentAuthorTobyBartels
• CommentTimeJul 23rd 2010

(By the way, […] Thoughts?)

I agree.

• CommentRowNumber18.
• CommentAuthorDavidRoberts
• CommentTimeJul 23rd 2010

supposed to be the 3-step factorization of a functor?

yes…. I thought the middle two categories weren’t equivalent, but now I see they are. hmm. I thought I’d correctly interpreted Toby’s notes to something I understand, but I’ll have another go at it.

• CommentRowNumber19.
• CommentAuthorMike Shulman
• CommentTimeJul 23rd 2010

The best definition I can think of for a 3-step factorization system is simply a pair of ordinary factorization systems $\left({E}_{1},{M}_{1}\right)$ and $\left({E}_{2},{M}_{2}\right)$ such that ${E}_{1}\subseteq {E}_{2}$ (and hence ${M}_{2}\subseteq {M}_{1}$). I think that condition implies that if you factor a morphism into $\left({E}_{1},{M}_{1}\right)$ and then factor the ${M}_{1}$ part into $\left({E}_{2},{M}_{2}\right)$, you get the same thing as if you first factored into $\left({E}_{2},{M}_{2}\right)$ and then factor the ${E}_{2}$ part into $\left({E}_{1},{M}_{1}\right)$, giving you the 3-step factorization. In the case of categories, ${E}_{1}=$ eso+full, ${M}_{1}$ = faithful, ${E}_{2}$ = eso, and ${M}_{2}$ = full+faithful. Probably for a (k+1)-step factorization system you similarly want k nested ordinary factorization systems.

• CommentRowNumber20.
• CommentAuthorTobyBartels
• CommentTimeJul 23rd 2010

Signs of run-away negative thinking: within a minute of reading Mike’s last comment, I decided that $n$-step factorisation systems make sense down through $n=-1$ in a way that fits in perfectly.

• CommentRowNumber21.
• CommentAuthorDavidRoberts
• CommentTimeJul 24th 2010

I figured out where I went wrong: instead to factoring the rightmost functor in

$D\to C\left[{D}_{0}\right]\to C$

I should factor the leftmost functor. Given a functor $f:A\to B$ which is the identity on objects, we can factor it as $A\to {\mathrm{im}}_{\mathrm{Mor}}f\to B$ where $\mathrm{Obj}\left({\mathrm{im}}_{\mathrm{Mor}}f\right)=\mathrm{Obj}\left(A\right)=\mathrm{Obj}\left(B\right)$ and $\mathrm{Mor}\left({\mathrm{im}}_{\mathrm{Mor}}f\right)$ is the image in $\mathrm{Mor}\left(B\right)$ of the arrow component of $f$. This fits the pattern I proposed better, in that the factorisation for 2-functors would then use the (epi,mono) factorisation in Set on 2-arrows, then the (eso,ff) factorisation in Cat on the level of 2- and 1-arrows, and then finally the (eso,local equivalence) factorisation in 2Cat.

In Toby’s polynomial notes the 3-step factorisation is not the same, but is equivalent to mine via the equivalence that Mike noted above, so at the very most we need uniqueness up to equivalence.

• CommentRowNumber22.
• CommentAuthorMike Shulman
• CommentTimeJul 24th 2010

I reorganized image along the lines I had in mind in comment 12. What should we do about coimage? I would kind of like to redirect it to image, so people get to the more general discussion; thoughts?

I would also like to propose that instead of “3-step” or “3-stage” factorization systems we use the word ternary, and similarly ordinary factorization systems are binary and general ones are k-ary. The prefixes “3-step” and “3-stage” confuse me as to whether they mean the factorization has three intermediate morphisms or three intermediate objects, but the words “ternary” and “binary” remind me that we are writing f as a ternary, resp. binary, composite of things, so that the number is the number of morphisms. Thoughts?

I can guess that a unary factorization system would be no structure at all (factor every morphism into a unary composite, namely itself). And that a nullary factorization system would factor every morphism into a composite of no things, i.e. an identity morphism – so that a category admits a nullary factorization system iff it is discrete. But I don’t know what a minus-unary factorization system would be; does it make the category empty? Contractible?

• CommentRowNumber23.
• CommentAuthorDavidRoberts
• CommentTimeJul 24th 2010

we use the word ternary, and similarly ordinary factorization systems are binary and general ones are k-ary

I agree.

As far as the definition of a k-ary factorisation system, is it possible to say the naive thing, and say that it is an ’section of k-ary composition’? Then a ’functorial k-ary factorisation’ is something like functorial binary factorisation being a section of (in the notation of Emily Riehl’s ’concise definition of a model cat’) the functor ${d}_{1}:{C}^{3}\to {C}^{3}$ – or is this too strong?

We could additionally require that for every partition $k=l+m$ the composites of the first $l$ and the last $m$ arrows gives us a binary factorisation. Or we could be unbiased and say that for every partition $k={m}_{1}+\dots +{m}_{j}$ we get a j-ary factorisation system by forming the ${m}_{i}$-ary composites… It would be nice to get this as a theorem from a basic description, though.

• CommentRowNumber24.
• CommentAuthorDavidRoberts
• CommentTimeJul 24th 2010
• (edited Jul 24th 2010)

(accidental copy of previous post)

• CommentRowNumber25.
• CommentAuthorTobyBartels
• CommentTimeJul 24th 2010
• (edited Jul 24th 2010)

@ Mike #22

I’d be inclined to say that ‘$n$-step’ implies $n$ morphisms while ‘$n$-stage’ implies $n$ objects. And actually, I’d prefer to count the objects! But I will adopt your terminology.

For $n>1$, I claim that an $n$-ary factorisation system consists of ${n}^{+}$ (that is $n+1$) factorisation systems $\left({E}_{i},{M}_{i}\right)$ (for $0\le i\le n$) such that

• ${M}_{i}\subseteq {M}_{{i}^{+}}$ for $0\le i (equivalently, ${E}_{i}\supseteq {E}_{{i}^{+}}$ for $0\le i),
• ${M}_{0}$ consists of only isomorphisms/equivalences (equivalently, ${E}_{0}$ consists of all morphisms), and
• ${M}_{n}$ consists of all morphisms (equivalently, ${E}_{n}$ consists of only isomorphisms/equivalences).

(Or course, an $n$-ary factorisation system is determined by the $n-1$ factorisations systems $\left({E}_{i},{M}_{i}\right)$ for $0, but the the other two exist.) Do you agree?

Given an $n$-ary factorisation system, the (co)image of $\left({E}_{i},{M}_{i}\right)$ is the $i$-(co)image of the entire $n$-ary factorisation system. (This agrees with the terminology in $\mathrm{Cat}$ for $n=3$, or more generally with the terminology in $\left(n-2\right)\mathrm{Cat}$ or even $\left(\infty ,n-2\right)\mathrm{Cat}$.)

Then extending this definition to lower values of $n$, every category (or $\infty$-category) has a unique $1$-ary factorisation system, where $\left({E}_{0},{M}_{0}\right)$ is (iso,all) and $\left({E}_{1},{M}_{1}\right)$ is (all,iso), as you suggested.

A category has a $0$-ary factorisation system if and only if it is a groupoid, in which case $\left({E}_{0},{M}_{0}\right)$ is both (iso,all) and (all,iso) at once. In other words, rather than requiring every morphism to be a $0$-ary composite on the nose, we require every morphism to be a $0$-ary composite up to isomorphism. I think that this is right, since a factorisation system (of any arity) should be given by specifying full and replete subcategories of the arrow category (or equivalently, collections of isomorphism classes of the arrow category), and every isomorphism is isomorphic to an identity.

I was wrong to say that $n=-1$ fit; the definition above does not actually make sense in that case, since the conditions required of the $0$ factorisation systems don’t parse. However, it seemed obvious to me that the empty category has a $\left(-1\right)$-ary factorisation systems, since we require an impossible condition of every morphism. But maybe that is a bad intuition.

Another thing that we should define is an $\infty$-ary factorisation system; $\infty \mathrm{Cat}$ has one of these. This consists of $\infty$ factorisation systems satisfying the first two conditions of $n$-ary factorisation systems, but there is no room for the last condition. Note that $\infty {\mathrm{Cat}}^{\mathrm{op}}$ has an ${\infty }^{\mathrm{op}}$-ary factorisation system.

• CommentRowNumber26.
• CommentAuthorMike Shulman
• CommentTimeJul 24th 2010

Do you agree?

Perhaps; that’s certainly a natural definition. I haven’t verified that it actually gives you a uniquely defined n-ary factorization of every morphism, as I have in the case n=3, but hopefully it’ll work out. That may be equivalent to David’s suggestion of giving a section of n-ary composition together with assuming that whenever $n=k+\ell$ we get a binary factorization system by composing the first $k$ and the last $\ell$ morphisms in the n-ary factorization.

I think I agree with your version of 0-ary factorizations as well. Not sure about (-1)-ary ones, though.

I also think this would be a good topic for a discussion at the Cafe. Perhaps I’ll start a post there.

• CommentRowNumber27.
• CommentAuthorTobyBartels
• CommentTimeJul 24th 2010

Perhaps I’ll start a post there.

If you do, please announce it here, since I don’t look at the Café much these days.

• CommentRowNumber28.
• CommentAuthorTobyBartels
• CommentTimeJul 24th 2010
• (edited Jul 24th 2010)

Another thing that we should define is an $\infty$-ary factorisation system […]

OK, I think that I’ve got that now. This also clarifies what an $\left(-1\right)$-ary factorisation system (called $0$-stage below) should be.

Fix any ordinal number (or opposite thereof, or any poset, really) $\alpha$. Then an $\alpha$-stage factorisation system (in an ambient $\infty$-category $C$) consists of an $\alpha$-indexed family of factorisation systems in $C$ such that:

• ${M}_{i}\subseteq {M}_{j}$ whenever $i\le j$ (equivalently, ${E}_{i}\supseteq {E}_{j}$ whenever $i\le j$),
• each morphism $f:X\to Y$ is both the inverse limit $\underset{i\to \infty }{\mathrm{lim}}{im}_{i}f$ in the slice category $C/Y$ and the direct limit $\underset{i\to -\infty }{colim}{coim}_{i}f$ in the coslice category $X/C$, and
• for each $f:X\to Y$, ${id}_{Y}$ is $\underset{i\to -\infty }{colim}{im}_{i}f$ and ${id}_{X}$ is $\underset{i\to \infty }{\mathrm{lim}}{coim}_{i}f$.

Results to check:

1. For every $\alpha$-indexed family of factorisation systems that satisfies the first of the three axioms above, there is a unique $\left(1+\alpha +1\right)$-stage factorisation system with the original factorisation systems in its middle, and every $\left(1+\alpha +1\right)$-stage factorisation system is of this form.
2. In particular, a $3$-stage factorisation system is equivalent to a factorisation system, and there is always a unique $2$-stage factorisation system.
3. A $1$-stage factorisation system exists iff $C$ is a groupoid, in which case it is unique.
4. An $n$-ary factorisation system from #25 is the same as an $\left(n+1\right)$-stage factorisation system.
5. A $0$-stage factorisation system exists iff $C$ is discrete, in which case it is unique.
6. An ${\alpha }^{\mathrm{op}}$-stage factorisation system in $C$ is the same as an $\alpha$-stage factorisation system in ${C}^{\mathrm{op}}$.
7. The factorisation systems of $\infty \mathrm{Cat}$ based on levels of fullness form an $\omega$-stage factorisation system in $\infty \mathrm{Cat}$.

The last result sets the numbering and also shows that we really do want examples where $\alpha$ is unbounded (important since the first result shows us how the definition can be made much simpler when $\alpha$ is bounded).

• CommentRowNumber29.
• CommentAuthorMike Shulman
• CommentTimeJul 28th 2010
• CommentRowNumber30.
• CommentAuthorTobyBartels
• CommentTimeJul 28th 2010

Thanks!