Want to take part in these discussions? Sign in if you have an account, or apply for one below
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
1 to 30 of 30
quick question, am in the middle of some computation:
I know that a “cotower diagram”
X0→X1→X2→⋯in some (cofibrantly generated) model category is projectively cofibrant precisely if each morphism is a cofibration and if the first object, hence all objects, are cofibrant.
Can we say something along these lines for general filtered diagrams?
(I guess it should be an easy generalization of the above statement, but I am throwing out this question anyway now.)
If I remember rightly the answer is YES. It involves the condition that each time you complete a pair of parallel morphisms by a further (co)refinement then the induced map from the (homotopy) pushout of the two maps to the new object is a cofibration.
I guess you mean the map out of the coproduct of the source with itself?
To lift in
A0→→A1↓↓B0→→B1↗↗X0→→X1we can proceed by first finding σ0 in
A0σ0↗↓X0→B0and then finding σ1:X1→A1 fitting into
A0∐A0→A1↓↓σ0∐σ0↗B0∐B0→B1↗↗X0∐X0→X1.So we’d need X0 to be cofibrant and X0∐X0→X1 to be a cofibration.
I’m not sure that that is the same as what I was meaning, but have to rush at the moment so will just say: if we have two maps from i to j then there is a k and a map from j to k equalising them (i.e. common corefinement)
This gives two maps from Xi to Xj and one from Xj to Xk. We need all the maps between things to be cofibrations but ALSO that the unique map induced from the pushout of the two maps with source Xi to the third object Xk, is also a cofibration.
Without spending time on doing the diagrams ….. I hope that is clearer. There may be further ’higher dimensional’ constraints but I don’t think so.
I did understand what you are saying about pushouts. I am just wondering if you really mean to say that.
Consider the “walking coequalizing” diagram
D={X0→→X1↘↙X2}.Then it seems to me that my previous message shows that in order for X•:D→C to be projectively cofibrant we need
X0 is cofibrant;
X0∐X0→X1 is a cofibration;
X1→X2 is a cofibration.
Of course pushout-morphisms being cofibrations becomes relevant for describing general cofibrations in the diagram category (as spelled out for example at Reedy model structure here, maybe that’s what you are thinking of?). But not for the characterization of cofibrant objects, I think.
If I remember rightly the answer is YES.
So is the statement you indicate is that for all filtered diagrams X we have: X is projectively cofibrant if it is a diagram of cofibrations (=all objects are cofibrant and all arrows are cofibrations)?
The argument given in Example 2.3.15, page 130 here works for any diagram X where each Xi has an incoming morphism from some other Xj for which a lifting has been shown.
Hi Stephan,
no, this can’t be right. Look at the example in #3, the simple case of a diagram consisting of two parallel morphisms. It is not sufficient that the two morphisms X0→X1 are cofibrations. You need X0∐X0→X1 to be a cofibration.
You can check that this is not only sufficient, which the argument in #3 shows, but also necessary:
for let
X0∐X0(a,b)→A1(f,g)↓↓pX1q→B1be a commuting diagram, where p:A1→B1 is an acyclic fibration. Using this we can build a lifting problem in the category of parallel morphisms, by setting
X0a→→bA1↓↓pX0p∘a=q∘f→→p∘b=q∘gB1↗↗qX0f→→gX1.Notice that the top square is an acyclic fibration in [=,𝒞]proj.
You see that a lift in the diagram category exists precisely if the original square has a lift, hence precisely if (f,g) is a cofibration. So for X0f→→gX1 to be projectively cofibrant, it is necessary that (f,g):X0∐X0→X1 is a cofibration.
the statement you indicate is that for all filtered diagrams X we have
no, this can’t be right
Yes, I thought it isn’t true that’s why I provoked this clear statement. Also the word ”filtered ” from the other thread is wrong here; e.g. a span is cofiltered but not filtered.
Okay, so if this is clear, here is an exercise:
what’s sufficient and necessary for a direct diagram to be projectively cofibrant?
For D a direct category this amounts to determine the cofibrant objects in the Reedy model structure where D=D+ and D− are the identities…I will try to dualize the exmple given there.
Right, for direct categories the projective and the Reedy structure coincide. You might even want to think about cofibrant diagrams in the Reedy model structure over any elegant Reedy category. (Mike will have some lemmas about that out, soon.)
But for this you have to do a little bit of genuine work. My “exercise” was meant as an immediate corollary, solvable essentially without work, just by looking at what we have so far.
As a warmup, consider a category with morphisms
0→→→1→→2and their composites (anything will do). So a category with three objects, with three morphisms from the 0th to the first, with two morphism from the second to the third, and with any choice of composition on these.
When is a diagram of this shape projectively cofibrant?
Hint: if this isn’t clear yet, try induction on n in the followig simple sense: write down a lifting problem. Then think about conditions you need to lift just over 0. After that, think about conditions to lift next just over 1. Use the previous discussion for that. After that, think about conditions to lift next over 2. Again use the previous discussion for that.
Once you have done this, you’ll see how it works generally for direct categories.
If you get stuck, you must let me know. Don’t hesitate to throw out questions.
Thanks, Urs!
As a warmup, consider a category with morphisms
I think for this example
0→→→1→→2the missing commuting diagrams (which look better in instiki than here) are
X0∐X0∐X0(a,b,c)→A1(f,g,h)↓↓pX1q→B1where p:A1→B1 is an acyclic fibration. And the corresponding lifting problem
X0a,b,c→A1↓↓pX0pa=qf,pb=qg,pc=qh→B1↗↗qX0f,g,h→X1.and hence the relevant necessary and sufficient conditions would be: (f1,g1):X1∐X1→X2 and (f,g,h)=(f0,g0,h0):X0∐X0∐X0→X1 ) shall be cofibrations. For a general set of n parallel morphisms we should require (pri=0,...,n+2):X×n+2n→Xn+1 to be a cofibration. This means that for any direct diagram to be projectively cofibrant it suffices indeed -as Tim says above- that it is a ”diagram of cofibrations” and all occurring (pri=0,...,n+2) are cofibrations.
Right, for direct categories the projective and the Reedy structure coincide. You might even want to think about cofibrant diagrams in the Reedy model structure over > any elegant Reedy category. (Mike will have some lemmas about that out, soon.)
Yes, if what I wrote above is correct so far, I would like to look into that, too.
Hi Stephan,
that looks right, though I thought you might want to give the argument for why this is the answer.
If you want to proceed to the general Reedy case, a next good exercise would be to look at the previous example again, by now formally computing the latching objects and the morphisms out of them and checking that this reproduces what you have above.
I thought you might want to give the argument for why this is the answer.
Before doing this I will write down what (I think) comes directly out of the definition of the latching objects:
As said above a direct category R becomes a Reedy category by choosing R=R+ and R−={r,idr|r∈R} (the discrete subcategory on all objects of R).
For X∈Fun(R,C) and r∈R we have by definition the latching object
LrX:=colims+→rXs:=colims∈R+/r−{idr}X(s)(the matching object MrX:=colims∈r/R−−{idr}X(s) we do not need this for the cofibrancy condition)
In the example
X0f0,g0,h0→X1f1,g1→X2f2→X3we have
L0X=colims∈R+/0−{id0}X(s)=0
L1X=colims∈R+/1−{id1}X(s)=coeq(f0,g0,h0)
L2X=colims∈R+/2−{id1}X(s)=coeq(f0,g0,h0)∐coeq(f1,g1)
L3X=colims∈R+/3−{id3}X(s)=coeq(f0,g0,h0)∐coeq(f1,g1)∐(X3,f2,id)
where the last cofactor is the couniversal cocone of the rightest subdiagram. For r∈R we have X is cofibrant iff LrX→Xr is a cofibration.
Now the statement from above giving necessary- and sufficient conditions for a diagram of two (respectively more) parallel morphisms to be Reedy-cofibrant should enter. For this we could write the colimit in question instead as the colimit of
X0i0→X0∐X0↓i1↓(f,g)X0∐X0(f,g)→X1(I have to interrupt now to catch a train)
Hi Stephan,
this is roughly going in the right direction, but needs some corrections: there are too many coeqs.
Let’s break this down in substeps and walk through it in small steps.
Consider just the category
R+:={0f→→g1}Then what is the category that we take the colimit over to produce the latching object at 1?
It is defined to be the full subcategory of the overcategory (R+)/1 on the non-identity elements.
So first of all, what is that overcategory (R+)/1? Its objects are the morphisms into “1”, so
Obj((R+)/1)={f,g,id1}.Its morphisms are the commuting triangles over “1”. There are precisely two non-identity triangles namely
0f→1f↘↙id11and0g→1g↘↙id11Agreed? So this means there are precisely two non-identity morphisms in (R+)/1, and so this category looks as follows:
(R+)/1={f→id1←g}.Okay?
But now the category we are after is supposed to be the full subcategory of this on all objects except id1. So we are to throw away that object together with all morphisms to and from it, but keep everything else. The result of that is
{f,g},which is the category with just two objects and no nontrivial morphisms.
Now for
X0→→X1a diagram in some model category 𝒞, the latching object at “1” is defined to be the colimit over the functor
{f,g}→𝒞which takes a morphism in R+ to the value Xdom of the above diagram at the domain of this morphism.
So this functor is the functor on the category with just two objects called f and g and no non-identity morphisms which assigns
f↦Xdom(f)=X0g↦Xdom(g)=X0Okay?
What is the colimit over such a diagram, which just consists of two copies of the object X0? It is the coproduct of these objects, hence
lim⟶({f,g}→𝒞)=X0∐X0.Hence this is the latching object L1X of our diagram.
Convince yourself that the canonical morphism L1X→X1 out of the latching object into X1 is
(f,g):X0∐X0→X1.So a necessary condition for Reedy cofibrancy = projective cofibrancy of the diagram X• is that this is a cofibration. (No coequalizers anywhere.)
Notice that this reproduces what we found before in #3! :-)
Let me know if this is clear so far. If not, let me know which step is unclear. If it is clear, redo the previous exercise! :-)
Another good exercise to do to help with the above, which you will need for lots of other things in homotopy theory anyway, is:
write down Δ+/[2].
Write down the full subcategory of it that is the shape for the colimit that gives the latching object at [2].
(Hint: it has six objects. A cartoon of it is depicted at Kan fibrant replacement.)
Hi Urs,
Let me know if this is clear so far
Yes it is. The error in my above writing was that I didn’t really exclude the identities.
Convince yourself that the canonical morphism L1X→X1 out of the latching object into X1 is (f,g):X0∐X0→X1.
This is clear by the universal property of the colimiting cocone since the canonical morphism can:X0∐X0→X1 satisfies can∘i1=f and can∘i2=g for the canonical inclusions i1,i2 and by the unicity of can and the fact that (f,g) solves this equalities.
redo the previous exercise! :-)
For more than two parallel morphisms your explanation with {f,g,h} instead of {f,g} applies mutatis mutandis and I guess in my above ”solution” the part where I combine the diagrams ”with the coproduct between the colimits” was right (by the universal properties of the colimits of the sub diagrams).
Another good exercise to do to help with the above, which you will need for lots of other things in homotopy theory anyway, is:
Yes, I will do this.
You might even want to think about cofibrant diagrams in the Reedy model structure over any elegant Reedy category. (Mike will have some lemmas about that out, soon.)
I suppose you think of a sequel of material in The univalence axiom for inverse diagrams?
Hi Stephan,
yes, okay. You might want to look at the diagram category
0→→1↘↙2just once more, to cross check your understanding (where the diagram is shorthand for saying that the composite of either of the two horizontal morphisms with the right one is equal to the left morphism). Here, the latching object at 2 does involve a coequalizer.
I suppose you think of a sequel of material in The univalence axiom for inverse diagrams?
Yes, there is a sequel in the making.
Yes, there is a sequel in the making.
But I’m not sure the relevance that it has here. Every presheaf on an elegant Reedy category (in a model category whose cofibrations are the monos) is Reedy cofibrant, since the Reedy cofibrations are then also the (levelwise) monos, but that’s already in Julie and Charles’ paper. What else are you thinking of?
Well, there is an op-ing here changing the variance. We have been discussing co-presheaves here.
yes, okay. You might want to look at the diagram category …
If we consider
0f,g→1↘h↙k2we have
R+/2={hf,g→k↘h↙kid2}
and a for a functor X of this shape the latching object over 2 gives me the opportunity to get the coequalizer off
L2X=colim(R+/2−{id2}X)→C)=colim({hf,g→k}X→C)=coeq(X(f),X(g))Then it seems to me that my previous message shows that in order for X•:D→C to be projectively cofibrant we need
X0 is cofibrant;
X0∐X0→X1 is a cofibration;
X1→X2 is a cofibration.
The latching object in 0 is the colimit over the empty diagram aka the initial object. This explains why X0 needs to be cofibrant. Condition 2 has already been clarified. It remains to see that X1→X2 being a cofibration implies that coeq(X(f),X(g))→X2 is, too (I do not see this instantly and will return to it later).
Another good exercise to do to help with the above, which you will need for lots of other things in homotopy theory anyway, is:
Let Δ+ denote the wide subcategory of Δ with only surjections as morphisms. These surjections are generated by cofaces only.
[0]d0,d1→[1]↘↙d0,d1,d2[2]The colimit over this diagram is the coproduct over the colimits of the sub diagrams consisting of just one morphism e.g. d0d0d0→d0 which are the cocones on the codomain e.g. (d0,idd0,d0d0d0→d0).
Well, there is an op-ing here changing the variance. We have been discussing co-presheaves here.
I thought the sequel will be (partly) about this op-ing?. Also in this thread we didn’t yet talk about an explicit choice of the model structure on the target category of the functor.
the sequel will be (partly) about this op-ing?
Yes. But, okay, I guess I misspoke somewhere above. The thing is:
we are currently looking at co-presheaf diagram on direct categories. This are presheaf diagrams in inverse categories. So that’s what Mike discussed in the earlier article.
My fault.
Hi Stephan,
re #24: Yes, you’ve got it! Good!
Indeed, the point you say you need to return to I also came across. I found it easiest to see by just writing down a lifting problem and checking that you can immediately find the required lifts that show this.
I thought the sequel will be (partly) about this op-ing?
No. The existing paper is about diagrams (or, if you must, “co-presheaves”) on inverse categories, which are presheaves on direct ones. The sequel is about presheaves on elegant Reedy categories, which generalize presheaves on direct ones.
Hi Urs,
Indeed, the point you say you need to return to I also came across. I found it easiest to see by just writing down a lifting problem and checking that you can immediately > find the required lifts that show this.
Maybe the two lifting problems are linked by the diagram with the sub diagrams
bottom:
X1→X2↘h↗coeqtop:
A→B↘id↗Asides:(edited)
Aid→Ap→B↑ϕh↑ϕ↑X1h→coeqp→X2and
Ap→B↑↑X1→X2where the letters with a hat are those induced from the ones without hat by the couniversal property of coeq. I will check tomorrow if this is correct.
Let p be an acyclic fibration giving a lifting problem of the putative cofibration coeq→X2 against an acyclic fibration p:A→B.
this gives a lifting problem
X1ϕh→A↓↓X2x→Bwhich possesses a solution l:X2→A by assumption and by commutativity of the 3d diagram this is also a solution of the original problem. (Note that it is easier to see this if we insert a identity cell X2→X2 instead of A→A in the 3d diagram - which I have edited since my post.)
1 to 30 of 30