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• CommentRowNumber1.
• CommentAuthorFosco
• CommentTimeApr 20th 2014

From theorem 1 proved in the page M-complete category follows that $({}^\perp M,M)$ is a factorization system. Obviously $M\subset ({}^\perp M)^\perp$: they are equal whenever $M$ is part of a prefactorization system. Can you give me an example of $M$ where this inclusion is strict?

• CommentRowNumber2.
• CommentAuthorMike Shulman
• CommentTimeApr 20th 2014

If you mean in the generality of “any class of maps $M$”, just consider $M=\emptyset$. Then $^\perp M$ is all maps and so $(^\perp M)^\perp$ is the isomorphisms.

• CommentRowNumber3.
• CommentAuthorFosco
• CommentTimeApr 20th 2014
• (edited Apr 20th 2014)

Thank you. I always forget trivial examples.

I have another problem with the proof of theorem 2; both the final step in the original proof in [CHK] and the sentence “Passing to adjuncts again, we find that $S g$ is also split epic” seem rather cryptic. Can you spend a word on it?

Trying to mesh the two arguments, I’m left with this verification: $S g$ is an isomorphism. It’s clear that $S g$ is a split mono. If I follow [CHK] I’m left to prove that $S g$ is a split epic; to do this, [CHK] builds the pullback diagram

$\begin{array}{ccc} Q &\overset{y}\to & T S A \\ {}^x\downarrow && \downarrow^{T S g} \\ X &\underset{\eta_X}\to &T S X \end{array}$

and then proves that $x$ is an isomorphism. Even if I take this for granted, how should I conclude that $S g$ admits a right inverse? If I pass to the adjunct of $T S g \circ y \circ x^{-1}=\eta_X$ I get $\epsilon_{S X} \circ S T S g \circ S y \circ S(x^{-1})=1$, from which the thesis follows when (for example) also $\epsilon_{S X}$ is invertible (by a retract-closure argument?).

• CommentRowNumber4.
• CommentAuthorFosco
• CommentTimeApr 20th 2014
• (edited Apr 20th 2014)

I am confused also by this:

$T S g$ is then also split monic, hence belongs to $M$ and thus also to $M'$.

This seems not to be a general result (I wouldn’t have any idea why it should be true). Instead, it seems to follow from the fact that $T(S g) \in T(\hom C)\subset ({}^\perp T(\hom C))^\perp = M_S$. Right?

((Edit: I added some details to your proof, in the hope of having made it slightly clearer for the occasional reader; I hope you won’t mind!))

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeApr 21st 2014

You seem to have answered our own questions. Although something seems to be missing from the second paragraph of your expanded proof.

• CommentRowNumber6.
• CommentAuthorFosco
• CommentTimeApr 21st 2014
• (edited Apr 21st 2014)

You’re right, there were a couple of typos. Now it works fine!

I wanted to check every detail to see where I precisely needed that the class $M$ in the definition of $M$-completeness is made by monomorphisms. Is there a similar notion of $M$-completeness for $\infty$-categories? Do I have to replace “monomorphisms” with something else, in that setting?

I want to compare the definition (“[…] We say that $C$ is $M$-complete if it admits all (even large) intersections of $M$-subobjects”) with Lurie’s HTT.6.1.6 which says “in the $\infty$-categorical context the emphasis on subobjects misses the point”. In particular, I can define subobjects in an $\infty$-category as equivalence classes of $(-1)$-truncated morphisms, but I don’t see why this should be the right point ov view to define $\infty$-$M$-completeness.

The question doesn’t seem trivial, as far as I can see, since the hierarchy between monic arrows becomes rather blurred in $\infty$-categorical setting:

In higher category theory, we may still consider the notion of “split monomorphism”, i.e. a morphism $m\colon A \to B$ in $C$ such that there exists a morphism $r\colon B \to A$ with $r \circ m$ being equivalent to the identity of $A$. However, in a higher category, such a morphism $m$ will not necessarily be a “monomorphism”, that is, it need not be $(-1)$-truncated.

In general […] in an (∞,1)-category, a split mono is not necessarily truncated at any finite level.

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeApr 21st 2014

An excellent question! I don’t know the answer.

• CommentRowNumber8.
• CommentAuthorFosco
• CommentTimeApr 22nd 2014
• (edited Apr 22nd 2014)

It seems that the only (or the main) problem is to figure out which is the right definition of monomorphism we want to adopt; let’s mantain it in a vague sense and backwards-engineer to see which is the feature of a monic arrow we need.

A quasicategory endowed with a distinguished class $\mathcal{K}$ of arrows is a marked simplicial set which happens to be a quasicategory. I denote with $\Mrk(C)$ the class of all markings of a fixed $\infty$-category $C$. This is a posetal class in the obvious sense.

Let now $\mathcal{K}$ be a marking of $C$ which is contained in the marking $Mono$ of monomorphisms of $C$. We say that $C$ is $\mathcal{K}$-complete if it admits all (even large) intersections of $\mathcal{K}$-objects, i.e. arbitrary (even large) limits of families $\{X_i\to A\}_{i\in J}$ for any fixed object $A\in C$.

More precisely, whenever we can consider a diagram $J\to \mathcal{K}$, where $J$ is possibly a class (depending on the ontology we want to consider, this can be either a (discrete, if we think simplicially) set outside of a universe $\mho$ fixed once and for all at the beginning of the discussion, or a proper class in $NBG$), then we can form the ($\infty$-)limit $lim \{X_i\to A\}_{i\in J}$ in the $\infty$-category of arrows $\mathcal{K}_{/A}\subseteq C_{/A}$.

This is a tentative to rephrase the first result:

Let $C$ be a quasicategory and $\mathcal{K}\in \Mrk(C)$ a marking such that

1. $C$ is $\mathcal{K}$-complete;
2. $\mathcal{K}$ is closed under composition, pullbacks and is contained in the marking $\Mono\in\Mrk(C)$.

Then the pair of markings $({}^\perp\mathcal{K},\mathcal{K})$ is a factorization system on $C$.

And this is a tentative rephrasing of the second:

Let $S\dashv T\colon A\leftrightarrows C$ an ($\infty$-)adjunction, $\mathcal{K}\in \Mrk(C)$ a class of monics which contains all the split monics. Let $A$ be $\mathcal{K}$-complete in the former sense. Then, if we consider the prefactorization $\mathbb{F}$ right-generated by $T(C_1)$ (i.e. $({}^\perp(T(C_1)), ({}^\perp(T(C_1)))^\perp)$, we have

1. ${}^\perp(T(C_1)) = \Sigma(S)$;
2. $\mathbb{F}$ is a factorization system.

(I’m being sloppy, but the things I didn’t define properly are easily adapted to the $\infty$-categorical setting.)

The proof of the first statement should go the same way than the classical one: consider an arrow $f\colon [1]\to C$ and the class $J_\mathcal{K}(f)$ of all objects in $C$ admitting a $\mathcal{K}$-arrow to the codomain of $f$, through which $f$ factors, i.e. all the commutative triangles $f=m p$, where $m\colon X\to B$ belongs to $\mathcal{K}$.

This can be seen as a diagram $J_\mathcal{K}(f)\to C^{[2]}$, whose composition with $d_0^*\colon C^{[2]}\to C^{[1]}$ gives a diagram $J_\mathcal{K}(f)\to\mathcal{K}$. Now, since $C$ is $\mathcal{K}$-complete, we can consider the wide pullback of all these arrows, and the universal property of this limit gives a factorization of $f$.

It remains to show that $p\in E={}^\perp\mathcal{K}$: to do this, we consider the same lifting problem, do the same pullback, and obtain the same factorization $p=n j$, with $j\in\mathcal{K}$. Now, $m n$ is a $\mathcal{K}$-arrow through which $f$ factors, hence there must be $b$ such that $m n b=m$; here the nature of the flavour in which we interpret the word “monomorphism” becomes important.

• CommentRowNumber9.
• CommentAuthorFosco
• CommentTimeApr 24th 2014

If you mean in the generality of “any class of maps $M$”, just consider $M=\emptyset$. Then $^\perp M$ is all maps and so $(^\perp M)^\perp$ is the isomorphisms.

Returning on this, I am confused (and I posed a confuesd question). If in the end we prove that $(E,M)$ is a factorization, it must in particular be a prefactorization, so that $M= E^\perp = ({}^\perp M)^\perp$. So what prevents me from having $M= ({}^\perp M)^\perp$ in general? $M=\varnothing$ obviously misses condition 3, like any other counterexample I can think about.

My original question should have been posed like this: there can’t be a $M$ satisfying 1,2,3,4 of Thm 1 without being equal to $({}^\perp M)^\perp$. So isn’t it true that I can prove WLOG the Corollary (which follows Thm 1), supposing that $M$+1,2,3,4 is part of a prefactorization system?

• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeApr 24th 2014

I don’t understand the question. If $M$ satisfies 1,2,3,4 of Theorem 1, then by Theorem 1 it is part of a factorization system, hence also part of a prefactorization system.

• CommentRowNumber11.
• CommentAuthorFosco
• CommentTimeApr 24th 2014
• (edited Apr 24th 2014)

My intuition finds rather surprising that 1,2,3,4 $\Rightarrow M=({}^\perp M)^\perp$; I think it’s me, I don’t see something obvious. So nevermind. Since you liked the question, I worked out some more details: I’m able to reproduce the results in that page if “monomorphism” is replaced with “monic arrow” in the sense of Joyal notes: an edge $f\colon [1]\to C$ is a monic in the q$\infty$-category $C$ if the square

$\begin{array}{ccc} S&\overset{1}\to & S\\ {}^1\downarrow && \downarrow^f\\ S&\underset{f}\to & T \end{array}$

is a homotopy limit. The next step is extend once more, $\infty$-izing the result in

D. Zangurashvili, Factorization systems and adjunctions, Georgian Mathematical Journal. Volume 6, Issue 2, Pages 191–200.

which says that

If $F\colon C\leftrightarrows X\colon U$ is a reflection and $X$ is $\mathcal{K}$-complete, then for any $(E,M)$ factorization system in $X$ we have a factorization system

$\Big(F^{-1}E^\circ, ({}^\perp U(M^\circ))^\perp \Big)$

[to be completed…]

• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeApr 25th 2014

Ah, so maybe the real question is, why in the definition of orthogonal factorization system does one not need to include the assumptions $M = (^\perp M)^\perp$ and $E = {}^\perp(E^\perp)$ (or equivalently that $E$ and $M$ are closed under retracts)? (E.g. they are omitted from the third definition given on the page orthogonal factorization system.) That is a bit surprising, and I don’t even remember off the top of my head why it’s true.

• CommentRowNumber13.
• CommentAuthorZhen Lin
• CommentTimeApr 25th 2014

There seems to be a typo in [CHK]. The functoriality of factorisations is implied by $E \subseteq {}^\bot M$ (or equivalently $M \subseteq E^\bot$), not the other inclusion. And once we have functorial unique factorisation, to determine whether a morphism is in $E$ (or $M$) it suffices to examine its functorial unique factorisation.

• CommentRowNumber14.
• CommentAuthorMike Shulman
• CommentTimeApr 25th 2014

A more symmetric notation for $E \subseteq {}^\bot M$ and $M \subseteq E^\bot$ is “$E\perp M$”.