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    • CommentRowNumber1.
    • CommentAuthorFosco
    • CommentTimeApr 20th 2014

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

    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeApr 20th 2014

    If you mean in the generality of “any class of maps MM”, just consider M=M=\emptyset. Then M^\perp M is all maps and so ( M) (^\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 SgS 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: SgS g is an isomorphism. It’s clear that SgS g is a split mono. If I follow [CHK] I’m left to prove that SgS g is a split epic; to do this, [CHK] builds the pullback diagram

    Q y TSA x TSg X η X TSX \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 xx is an isomorphism. Even if I take this for granted, how should I conclude that SgS g admits a right inverse? If I pass to the adjunct of TSgyx 1=η XT S g \circ y \circ x^{-1}=\eta_X I get ε SXSTSgSyS(x 1)=1\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 ε SX\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:

    TSgT S g is then also split monic, hence belongs to MM and thus also to MM'.

    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(Sg)T(homC)( T(homC)) =M ST(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 MM in the definition of MM-completeness is made by monomorphisms. Is there a similar notion of MM-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 CC is MM-complete if it admits all (even large) intersections of MM-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)(-1)-truncated morphisms, but I don’t see why this should be the right point ov view to define \infty-MM-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:ABm\colon A \to B in CC such that there exists a morphism r:BAr\colon B \to A with rmr \circ m being equivalent to the identity of AA. However, in a higher category, such a morphism mm will not necessarily be a “monomorphism”, that is, it need not be (1)(-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 𝒦\mathcal{K} of arrows is a marked simplicial set which happens to be a quasicategory. I denote with Mrk(C)\Mrk(C) the class of all markings of a fixed \infty-category CC. This is a posetal class in the obvious sense.

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

    More precisely, whenever we can consider a diagram J𝒦J\to \mathcal{K}, where JJ 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 NBGNBG), then we can form the (\infty-)limit lim{X iA} iJlim \{X_i\to A\}_{i\in J} in the \infty-category of arrows 𝒦 /AC /A\mathcal{K}_{/A}\subseteq C_{/A}.

    This is a tentative to rephrase the first result:

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

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

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

    And this is a tentative rephrasing of the second:

    Let ST:ACS\dashv T\colon A\leftrightarrows C an (\infty-)adjunction, 𝒦Mrk(C)\mathcal{K}\in \Mrk(C) a class of monics which contains all the split monics. Let AA be 𝒦\mathcal{K}-complete in the former sense. Then, if we consider the prefactorization 𝔽\mathbb{F} right-generated by T(C 1)T(C_1) (i.e. ( (T(C 1)),( (T(C 1))) )({}^\perp(T(C_1)), ({}^\perp(T(C_1)))^\perp), we have

    1. (T(C 1))=Σ(S){}^\perp(T(C_1)) = \Sigma(S);
    2. 𝔽\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:[1]Cf\colon [1]\to C and the class J 𝒦(f)J_\mathcal{K}(f) of all objects in CC admitting a 𝒦\mathcal{K}-arrow to the codomain of ff, through which ff factors, i.e. all the commutative triangles f=mpf=m p, where m:XBm\colon X\to B belongs to 𝒦\mathcal{K}.

    This can be seen as a diagram J 𝒦(f)C [2]J_\mathcal{K}(f)\to C^{[2]}, whose composition with d 0 *:C [2]C [1]d_0^*\colon C^{[2]}\to C^{[1]} gives a diagram J 𝒦(f)𝒦J_\mathcal{K}(f)\to\mathcal{K}. Now, since CC is 𝒦\mathcal{K}-complete, we can consider the wide pullback of all these arrows, and the universal property of this limit gives a factorization of ff.

    It remains to show that pE= 𝒦 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=njp=n j, with j𝒦j\in\mathcal{K}. Now, mnm n is a 𝒦\mathcal{K}-arrow through which ff factors, hence there must be bb such that mnb=mm 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 MM”, just consider M=M=\emptyset. Then M^\perp M is all maps and so ( M) (^\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)(E,M) is a factorization, it must in particular be a prefactorization, so that M=E =( M) M= E^\perp = ({}^\perp M)^\perp. So what prevents me from having M=( M) M= ({}^\perp M)^\perp in general? M=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 MM satisfying 1,2,3,4 of Thm 1 without being equal to ( M) ({}^\perp M)^\perp. So isn’t it true that I can prove WLOG the Corollary (which follows Thm 1), supposing that MM+1,2,3,4 is part of a prefactorization system?

    • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeApr 24th 2014

    I don’t understand the question. If MM 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 M=( M) \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:[1]Cf\colon [1]\to C is a monic in the q\infty-category CC if the square

    S 1 S 1 f S f T \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:CX:UF\colon C\leftrightarrows X\colon U is a reflection and XX is 𝒦\mathcal{K}-complete, then for any (E,M)(E,M) factorization system in XX we have a factorization system

    (F 1E ,( U(M )) ) \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=( M) M = (^\perp M)^\perp and E= (E )E = {}^\perp(E^\perp) (or equivalently that EE and MM 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 ME \subseteq {}^\bot M (or equivalently ME M \subseteq E^\bot), not the other inclusion. And once we have functorial unique factorisation, to determine whether a morphism is in EE (or MM) it suffices to examine its functorial unique factorisation.

    • CommentRowNumber14.
    • CommentAuthorMike Shulman
    • CommentTimeApr 25th 2014

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