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In Emily Riehl's Categorical Homotopy Theory, there is a section on Garner's Small Object Argument which I'm trying and failing to understand. Originally I followed most of Garner's paper, using the book to try and explicitly understand the stages in the transfinite construction.

I guess I'll try to sketch some details first. Garner's small object argument is a a refinement of Quillen's in that it is universal and converges. The point is in to construct a 'free natural weak factorization system', i.e in reflecting along a semantics functor $\mathcal G:\mathsf{nwfs}(\mathsf{C})\longrightarrow \mathsf{Cat}/\mathsf{C^2}$. This problem is simplified by factoring $\mathcal G$ in a reasonably canonical way, because abstract nonsense gives two of the three reflections needed. The final reflection is then shown to be equivalent to constructing a free monoid, which is then shown to be equivalent to the convergence of a certain transfinite construction called the 'free monoid sequence'.

(Now we come to proposition 4.22 in Garner's paper.)

A convergence criterion in the context of our problem is the condition the functor $R=d_0\circ F:\mathsf{C^2}\longrightarrow \mathsf{C^2}$ preserves $\lambda$-filtered colimits, where $d_0$ is the image of the simplicial $\delta_0$, sending a composite of arrows to the first one. By abstract nonsense again it suffices to prove $L=d_2\circ F$ preserves such colimits. By abstract nonsense, $L$ admits the following explicit description. First, given the object $I:\mathsf{J}\rightarrow \mathsf{C^2}$ we want to reflect, form its left Kan extension along itself $M=\operatorname{Lan}_II:\mathsf{C^2}\longrightarrow \mathsf{C^2}$.

Remark 1. Remark 12.5.2 in the book says that left Kan extending a functor along itself yields (by abstract nonsense) a comonad, called the density comonad.

With the usual coend formula, the component at $f$ of the counit $\epsilon :M\Rightarrow 1_{\mathsf{C^2}}$ is the arrow $$\int^{j\in \mathsf J}\mathsf{C^2}(Ij,f)\circ Ij\longmapsto f$$ which is adjunt to the identity natural transformation on $\mathsf{C^2}(I(-),f)$. This $\epsilon$ can then be factored as the composite $M\overset{\xi}{\Longrightarrow}L\overset{\Phi}{\Longrightarrow}1_{\mathsf{C^2}}$ where the components of $\xi$ are pushouts and the components of $\Phi$ are identities on its domain.

By abstract nonsense, $L$ preserves any colimit $M$ does, so we need to find an ordinal $\lambda$ for $M$. The proof ends by using the smallness condition to interchange some colimits. $\blacksquare$

It looks like section 12.4 of the book explain the process a little, but not enough for me. Suppose we want to factor an arrow $f$. The author says the square (12.2.3) should be thought of as a "generic lifting problem" that tests whether or not $f$ is a fibration, which makes sense. The fill in $\phi_f$ in (12.4.1) is referred to as a lifting function, which also makes sense. In step-one we factor the above square through the pushout of its cospan. This yields an equivalent lifting problem, proving that $f\in \mathcal J^\pitchfork$ iff it has has the right lifting property against $Lf$ in the canonical square (triangle) $Rf\circ Lf=f\circ 1$. We iterate this process, modding out redundancies, to ensure (using smallness) that it converges.

Remark 2. If I understand correctly, Garner says $\epsilon$ is component-wise exactly the transition to the "lifting function" point of view. I am confused about the second sentence regarding the equivalent lifting problem...

I would like to understand what the density comonad is (preferably at a level of generality where I can make out the relevance to this context), why, conceptually, does it encode the passage to an equivalent (but much more canonical) lifting problem, and whether the quotienting described in section 6.5 of Garner's paper, which gives convergence, is also encoded in it.

Finally, let me emphasize I did not learn Quillen's small object argument before, so answers like "this is just a modification of the standard argument" will not help me very much. I am trying to learn this "corrected" version from scratch. I have also never met density comonads or codensity monads before.

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  • $\begingroup$ I'm having a hard time understanding exactly what you are asking. Is part of the question what density comonads have to do with lifting problems? If so, maybe it helps to say that when $\mathsf{J}$ is discrete, the component at $f$ of the counit of $M$ is exactly the generic lifting problem (12.2.3). (But I get the feeling you understood that and wanted something else?) $\endgroup$ – Omar Antolín-Camarena May 1 '16 at 18:57
  • $\begingroup$ In your Remark 2, what "second sentence" are you referring to? $\endgroup$ – Omar Antolín-Camarena May 1 '16 at 18:58
  • $\begingroup$ @OmarAntolín-Camarena sorry for the late reply. I would like some help in understanding what the density comonad does. Conceptually, why does its counit describe generic lifting problems, and what does it do for more interesting $\mathsf J$? What does it measure about the functor $I$? $\endgroup$ – Arrow May 3 '16 at 8:19
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    $\begingroup$ If you'd like a general introduction to codensity monads, there's this: golem.ph.utexas.edu/category/2012/09/… . If you'd like an introduction to density comonads, I'm afraid I can't help :-) $\endgroup$ – Tom Leinster May 3 '16 at 19:31
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    $\begingroup$ Or maybe I can. Here's how to think about the density comonad of a functor $F$: it's what the comonad induced by $F$ and its right adjoint would be if $F$ had a right adjoint - but it's defined in many situations where $F$ doesn't have a right adjoint. $\endgroup$ – Tom Leinster May 3 '16 at 19:33
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Maybe it would help to have:

  1. A definition of "$f \in \mathsf{J}^\pitchfork$" when $\mathsf{J}$ is a category equipped with a functor $I : \mathsf{J} \to \mathsf{C}^2$; a definition that does not mention the density comonad of $I$.

  2. An argument showing that $f \in \mathsf{J}^\pitchfork$ is equivalent to having a diagonal filler for the counit $\mathrm{Lan}_II(f) \to f$.

[In the case that $\mathsf{J}$ is discrete the definition in step 1 would be simply that for each $j \in \mathsf{J}$ and each square $j \to f$ there is a diagonal filler. And in step 2 you'd argue that the counit $\mathrm{Lan}_II(f) \to f$ is exactly the square (12.2.3), and that having a diagonal filler for it provides all at once the required diagonals for all the individual squares $j \to f$.]

OK, the definition of $f \in \mathsf{J}^\pitchfork$ is that you can choose for each object $j \in \mathsf{J}$ and each square $\alpha : I(j) \to f$ a diagonal filler $\phi_\alpha$ in such a way that the fillers are compatible with composition in $\mathsf{J}$, namely, given a morphism $g : j \to j'$ and a square $\alpha' : I(j') \to f$ the fillers for $\alpha'$ and $\alpha = \alpha' \circ I(g)$ satisfy $\phi_\alpha = \phi_{\alpha'} \circ I_{\mathsf{cod}}(g)$ (where (1) $I_\mathsf{cod}(g)$ means the bottom arrow of the square $I(g) : I(j) \to I(j')$ --bottom if you draw $I(j)$ going down-- and (2) I apologize for not knowing how to draw diagrams with diagonal arrows on MO).

Now, how do we see that giving these compatible fillers $\phi$ is the same as filling the counit $\mathrm{Lan}_II(f) \to f$? Well, first notice the compatibility with composition of the $\phi$'s can be rephrased as saying that $\phi$ gives a functor from the comma category $I \downarrow f$ to $\mathsf{C}^2$. In fact, the collection of lifting problems $I(j) \to f$ form a cocone with vertex $f$ over the diagram $I \downarrow f \xrightarrow{\pi} \mathsf{J} \xrightarrow{I} \mathsf{C}^2$. And now just recall the colimit formula for the left Kan extension: $\mathrm{Lan}_II(f) = \mathrm{colim}(I \downarrow f \xrightarrow{\pi} \mathsf{J} \xrightarrow{I} \mathsf{C}^2)$. The counit of the density comonad is corresponds to the cocone mentioned above.

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  • $\begingroup$ Super clear! I will read carefully later, but I think your answer resolves my issue! $\endgroup$ – Arrow May 4 '16 at 8:44
  • $\begingroup$ Just a question - why do you take the codomain of the functor $\phi$ to be $\mathsf{C^2}$ instead of $I\downarrow \operatorname{dom}f$? Is it just to make all the compositions defined? $\endgroup$ – Arrow May 4 '16 at 12:43
  • $\begingroup$ Well, I just wrote $\mathsf{C}^2$ out of laziness (plus what I wanted to emphasize is the domain: $\mathsf{I} \downarrow f$), you can certainly be more precise. I didn't list the full set of requirements for $\phi$ in that last paragraph, for example, you still have to ask that each individual $\phi_\alpha$ actually solve its lifting problem! $\endgroup$ – Omar Antolín-Camarena May 4 '16 at 16:55
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    $\begingroup$ Oh, and I don't think $I \downarrow \mathrm{dom} f$ is quite right, @Arrow: the codomain of $I$ is $\mathsf{C}^2$ but $\mathrm{dom} f \in \mathsf{C}$. You could say $\phi : I \downarrow F \to I_\mathsf{cod} \downarrow \mathrm{dom} f$, where $I_\mathsf{cod}$ is again the composite $\mathsf{J} \xrightarrow{I} \mathsf{C}^2 \xrightarrow{\mathrm{cod}} \mathsf{C}$. If you add that $\phi$ commutes with the projections to $\mathsf{J}$, that captures all the proper domain and codomain information. (You still have to ask that the triangles in the indvidual lifting problems commute!) $\endgroup$ – Omar Antolín-Camarena May 4 '16 at 17:01

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