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1. • CommentRowNumber1.
   • CommentAuthorFosco
   • CommentTimeApr 20th 2014
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   Author: Fosco
   Format: MarkdownItexFrom theorem 1 proved in the page [M-complete category](http://ncatlab.org/nlab/show/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?

   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?

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeApr 20th 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexIf 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.

   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.

3. • CommentRowNumber3.
   • CommentAuthorFosco
   • CommentTimeApr 20th 2014
   • (edited Apr 20th 2014)
   • PermaLink
   Author: Fosco
   Format: MarkdownItexThank 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?).

   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?).

4. • CommentRowNumber4.
   • CommentAuthorFosco
   • CommentTimeApr 20th 2014
   • (edited Apr 20th 2014)
   • PermaLink
   Author: Fosco
   Format: MarkdownItexI 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!))

   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!))

5. • CommentRowNumber5.
   • CommentAuthorMike Shulman
   • CommentTimeApr 21st 2014
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   Author: Mike Shulman
   Format: MarkdownItexYou seem to have answered our own questions. Although something seems to be missing from the second paragraph of your expanded proof.

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

6. • CommentRowNumber6.
   • CommentAuthorFosco
   • CommentTimeApr 21st 2014
   • (edited Apr 21st 2014)
   • PermaLink
   Author: Fosco
   Format: MarkdownItexYou'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 [[equivalence|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 object|truncated]]. > > In general [...] in an [[(∞,1)-category]], a split mono is not necessarily truncated at any finite level.

   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.

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeApr 21st 2014
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   Author: Mike Shulman
   Format: MarkdownItexAn excellent question! I don't know the answer.

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

8. • CommentRowNumber8.
   • CommentAuthorFosco
   • CommentTimeApr 22nd 2014
   • (edited Apr 22nd 2014)
   • PermaLink
   Author: Fosco
   Format: MarkdownItexIt 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.

   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.

9. • CommentRowNumber9.
   • CommentAuthorFosco
   • CommentTimeApr 24th 2014
   • PermaLink
   Author: Fosco
   Format: MarkdownItex> 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 *pre*factorization, 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?

   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?

10. • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeApr 24th 2014
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexI 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.

    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.

11. • CommentRowNumber11.
    • CommentAuthorFosco
    • CommentTimeApr 24th 2014
    • (edited Apr 24th 2014)
    • PermaLink
    Author: Fosco
    Format: MarkdownItexMy 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...]

    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…]

12. • CommentRowNumber12.
    • CommentAuthorMike Shulman
    • CommentTimeApr 25th 2014
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexAh, 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.

    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.

13. • CommentRowNumber13.
    • CommentAuthorZhen Lin
    • CommentTimeApr 25th 2014
    • PermaLink
    Author: Zhen Lin
    Format: MarkdownItexThere 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.

    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.

14. • CommentRowNumber14.
    • CommentAuthorMike Shulman
    • CommentTimeApr 25th 2014
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexA more symmetric notation for $E \subseteq {}^\bot M$ and $M \subseteq E^\bot$ is "$E\perp M$".

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