better then deleting ill-placed information is to give it its proper place. So I have re-instantiated and expanded the pointer in the line before (now here).

]]>Removed “replete image” from the example of images in Cat. The essential image is the image wrt replete subcategories.

]]>Removing the redirect corestriction and starting a separate page for it.

]]>Removing corestriction and starting a separate page for it.

]]>I added a parenthetical “($M$)-” to make the context clearer.

I looked at the parenthetical “(i.e., $f = m e$)” and I personally find it unobtrusive. In addition, it may well be that the author of that parenthetical has experience of other readers who would actually find that helpful. Let’s leave it be.

Edit: I also made a footnote at extremal epimorphism, clarifying which notion of image is meant.

]]>I think you have introduced confusion in the article image by not paying close contextual attention to the fact that it had been written for a *general* notion of ($M$)-image where $M$ is a given class of monomorphisms. That is, extremal monomorphism makes reference just to the default meaning of “monomorphism”. I’m going to roll back to version #39 (edit: #40, I mean) and see if appropriate changes are nevertheless warranted.

It also seems to me the proviso “has equalizers” is appropriate. I’m not sure why you think you speak for “readers” generally. If you have questions about the mathematics of an article, then please ask.

]]>image did not mention the concept of extremal epimorphism until about midway into the article. In contrast, extremal epimorphism says

An image factorization of a morphism $f$ is,

by definition, a factorization $f = m \circ e$ where $m$ is a monomorphism and $e$ is an extremal epimorphism.

where the boldface has been introduced for this comment only.

Readers might therefore expect extremal epimorphism to already appear within section “2. Definition” of image and be a bit puzzled that it does not.

Therefore added a reference to [extremal epimorphism, mediated by some slight reformulation.

Incidentally, removed the parenthetical “(then $f = m e$)” which while maybe useful to remind readers of “Leibnizian” order of composition, seems too much here.

Readers’ experience remains slightly more rough than necessary, on account of the proviso “if […] C$ has equalizers” in the section Relation to factorization systems).

]]>The pages *image* and *(epi,mono) factorization system* and also *abelian category* knew nothing of each other. I have added some cross-links.

Thanks!

]]>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 $(-1)$-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 $i \leq j$ (equivalently, $E_i \supseteq E_j$ whenever $i \leq j$),
- each morphism $f\colon X \to Y$ is both the inverse limit $\underset{i \to \infty}\lim \im_i f$ 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 $f\colon X \to Y$, $\id_Y$ is $\underset{i \to -\infty}\colim \im_i f$ and $\id_X$ is $\underset{i \to \infty}\lim \coim_i f$.

Results to check:

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

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).

]]>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.

]]>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.

]]>@ 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 $n \gt 1$, I claim that an **$n$-ary factorisation system** consists of $n^+$ (that is $n + 1$) factorisation systems $(E_i,M_i)$ (for $0 \leq i \leq n$) such that

- $M_i \subseteq M_{i^+}$ for $0 \leq i \lt n$ (equivalently, $E_i \supseteq E_{i^+}$ for $0 \leq i \lt n$),
- $M_0$ 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 $(E_i,M_i)$ for $0 \lt i \lt n$, but the the other two exist.) Do you agree?

Given an $n$-ary factorisation system, the (co)image of $(E_i,M_i)$ is the **$i$-(co)image** of the entire $n$-ary factorisation system. (This agrees with the terminology in $Cat$ for $n = 3$, or more generally with the terminology in $(n - 2) Cat$ or even $(\infty,n - 2) Cat$.)

Then extending this definition to lower values of $n$, every category (or $\infty$-category) has a unique $1$-ary factorisation system, where $(E_0,M_0)$ is (iso,all) and $(E_1,M_1)$ 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 $(E_0,M_0)$ 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 $(-1)$-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; $\infty Cat$ 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 $\infty Cat^op$ has an $\infty^op$-ary factorisation system.

]]>(accidental copy of previous post)

]]>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 $d_1:C^\mathbf{3} \to C^\mathbf{3}$ – 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 $k=m_1 + \ldots + m_j$ 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.

]]>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?

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

$D \to C[D_0] \to C$I should factor the leftmost functor. Given a functor $f:A \to B$ which is the identity on objects, we can factor it as $A \to im_{Mor} f \to B$ where $Obj(im_{Mor} f) = Obj(A)=Obj(B)$ and $Mor(im_{Mor} f)$ is the image in $Mor(B)$ 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.

]]>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.

]]>The best definition I can think of for a 3-step factorization system is simply a pair of ordinary factorization systems $(E_1,M_1)$ and $(E_2,M_2)$ 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 $(E_1,M_1)$ and then factor the $M_1$ part into $(E_2,M_2)$, you get the same thing as if you first factored into $(E_2,M_2)$ and then factor the $E_2$ part into $(E_1,M_1)$, 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.

]]>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.

]]>(By the way, […] Thoughts?)

I agree.

]]>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).

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

I was wondering about this a few times, but never did anything.

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.

]]>