Timeline for When does the forgetful functor from algebras over a monad commute with homotopy geometric realizations?
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| Jun 1, 2018 at 23:37 | comment | added | Hadrian Heine | Of course $dom(\theta) \circ q_1 $ and $dom(\theta) \circ q_2$ are both lifts in the same square $\theta \circ \alpha_1 = \theta \circ \alpha_2$. But lifts are not unique. So my question is how to find the lift $q$? | |
| Jun 1, 2018 at 23:34 | comment | added | Hadrian Heine | If $q$ is a lift in the square $\alpha: p \to Hom(Y,F)$, then $q_1$ is a lift in the square $\alpha_1$ and $q_2$ a lift in the square $ \alpha_2 $ and we have $dom(\theta) \circ q_1 = dom(\theta) \circ q_2$. But if we choose $q_1$ to be a lift in the square $\alpha_1: p \to F_2$ and $q_2$ to be a lift in the square $\alpha_2: p \to F_2,$ how do we get that we have $dom(\theta) \circ q_1 = dom(\theta) \circ q_2.$ | |
| Jun 1, 2018 at 23:17 | comment | added | Hadrian Heine | So as $\alpha_1 \neq \alpha_2 $ in general, especially $pr_1 \circ q \neq pr_2 \circ q$ in general. i.e. $q =(q_1, q_2)$ for some maps $q_1, q_2: A' \to dom( F_2) $ with $q_1 \neq q_2$ in general and $dom(\theta) \circ q_1 = dom(\theta) \circ q_2$ and $\theta: F_2 \to F_1$ the canonical morphism from above. How do we get $q=(q_1,q_2)?$ | |
| Jun 1, 2018 at 23:12 | comment | added | Hadrian Heine | Ok. But my question was the following concerning the example, where $Y= \Delta^1 \times \Delta^1$: Let's given a square $ \alpha : p \to Hom(Y, F) = F_2 \times_{F_1} F_2 $ with $\alpha_1:= pr_1 \circ \alpha$ and $\alpha_2:= pr_2 \circ \alpha$ as defined above and $p: A \to A'$ a cofibration. Let $ q: A' \to dom( Hom(Y, F) ) = dom(F_2 \times_{F_1} F_2 ) $ a lift in the square $\alpha.$ Then $pr_1 \circ q \circ p= \alpha_1 $ and $pr_2 \circ q \circ p= \alpha_2. $ | |
| Jun 1, 2018 at 21:53 | comment | added | Dmitri Pavlov | One can also rephrase the above argument as follows: the commutativity of two triangles in the square α: p→Hom(Y,F) is equivalent by the universal property of equalizers to the commutativity of two triangles in the square p→∏F(Δ^N). The latter holds by construction of the original map A'→dom(∏F(Δ^N)) via the lifting property. | |
| Jun 1, 2018 at 21:50 | comment | added | Dmitri Pavlov | For the bottom triangle we compose with the map codom(Hom(Y,F))→codom(∏F(Δ^N)), and use the fact that the original map A'→dom(∏F(Δ^N)) makes this triangle commute. | |
| Jun 1, 2018 at 21:48 | comment | added | Dmitri Pavlov | @HadrianHeine: Sorry, I misunderstood your notation. Of course, α_1 is not α_2. What I meant to say is that the map A'→dom(∏F(Δ^N)) is constructed first. We then construct a map A'→dom(Hom(Y,F)) by verifying that the two compositions A'→dom(∏F(Δ^N))→dom(∏F(Δ^{N−1})) agree, which is explained in my May 25 comments. We have constructed the diagonal in the square α: p→Hom(Y,F). We now verify that the two triangles commute. For the upper triangle we compose with the map dom(Hom(Y,F))→dom(∏F(Δ^N)) and use the fact that the original map A'→dom(∏F(Δ^N)) makes this triangle commute. | |
| Jun 1, 2018 at 17:12 | comment | added | Hadrian Heine | Why is $\alpha_1 = \alpha_2$ in the example? $\alpha : p \to Hom(Y, F) = F_2 \times_{F_1} F_2 $ is a square in $C$ so that we have $\theta \circ \alpha_1 = \theta \circ \alpha_2$ if $\theta: F_2 \to F_1$ denotes the morphism induced by the map $d_1: \Delta^1 \to \Delta^2.$ | |
| May 31, 2018 at 23:02 | comment | added | Dmitri Pavlov | @HadrianHeine: The only time the lifting property is used is when we construct the map q. The other maps are constructed using the universal property of equalizers. (My usage of the word "lift" was ambiguous above.) In particular, α=γ and q=q' by definition. Also α_1=α_2 by definition (for the example). In other words, the map q:A'→dom(F(Δ^N)) is constructed first. We then verify that q factors through the equalizer by verifying that the two compositions A'→dom(F(Δ^{N−1})) agree. This yields a map A'→eq. Finally, we verify that the two triangles commute using the universal property. | |
| May 30, 2018 at 0:09 | comment | added | Hadrian Heine | We get a lift $ q: A' \to dom(F_2)$ in the square $\alpha_1: p \to Hom(Y, F) = F_2 \times_{F_1} F_2 \to F_2.$ Why is $q$ also a lift in the quare $\alpha_2: p \to Hom(Y, F) = F_2 \times_{F_1} F_2 \to F_2.$ The map $q: A' \to dom(F_2)$ is a lift in the square $(\alpha_1, \alpha_1): p \to Hom(Y, F) = F_2 \times_{F_1} F_2 .$ | |
| May 30, 2018 at 0:05 | comment | added | Hadrian Heine | Applied to the example $Y = \Delta^1 \times \Delta^1$: Let $A \to A'$ a cofibration of $C$ and $\alpha : p \to Hom(Y, F) = F_2 \times_{F_1} F_2$ a square in $C.$ Denote $pr_1, pr_2: Hom(Y, F) = F_2 \times_{F_1} F_2 \to F_2$ the first and second projection induced by the two maps $\Delta^2 \to \Delta^1 \times \Delta^1, 0 \mapsto (0,0), 1 \mapsto (0,1), 2 \mapsto (1,1)$ respectively $0 \mapsto (0,0), 1 \mapsto (1,0), 2 \mapsto (1,1).$ Set $\alpha_1 := pr_1 \circ \alpha, \alpha_2:= pr_2 \circ \alpha. $ | |
| May 29, 2018 at 23:12 | comment | added | Hadrian Heine | Can we choose N such that $\alpha$ equals $\gamma$ to produce a lift in the square $\alpha$? | |
| May 29, 2018 at 23:05 | comment | added | Hadrian Heine | Let $p: A \to A'$ be a cofibration in C. Let a square $\alpha: p \to Hom(Y,F)$ be given. We get a lift q:A'→dom(F(Δ^N)) in the square $\beta: p \to Hom(Y,F) \to Hom(\Delta^N,F)$.The diagonal map $ \bar{q}:A'→dom(∏_i F(Δ^N)) $ is a lift in the diagonal square $\bar{\beta}: p \to ∏_i F(Δ^N) $. $ \bar{\beta}$ lifts to a square $\gamma: p \to Hom(Y,F)$ and $ \bar{q}:A'→dom(∏_i F(Δ^N)) $ lifts to a map $ q': A'→dom(F(Y)) $ that is a lift in the square $\gamma: p \to Hom(Y,F)$. | |
| May 25, 2018 at 22:49 | comment | added | Dmitri Pavlov | @HadrianHeine: We now construct a lift of a generating cofibration p:A→A' with respect to Hom(Y,F). We lift p against F(Δ^N), obtaining a map q:A'→dom(F(Δ^N)), hence also the diagonal map q:A'→dom(∏_i F(Δ^N)), which lifts to a map A'→dom(F(Y)) by the universal property of the above coequalizer. | |
| May 25, 2018 at 22:49 | comment | added | Dmitri Pavlov | @HadrianHeine: Likewise, j runs over all interior nondegenerate (N−1)-simplices of Y, which can be identified with N-tuples of vertices such that each k-tuple but one is obtained from the previous k-tuple by increasing exactly one component by 1, and one k-tuple is obtained from the previous k-tuples by increasing exactly two different components a and b by 1. For each j there are exactly two different i's that have j as a face, obtained by inserting one intermediate tuple that increases either a or b by 1. Observe that the face number for both such i's is the same. | |
| May 25, 2018 at 22:49 | comment | added | Dmitri Pavlov | @HadrianHeine: Yes. This can be seen in complete generality as follows. Suppose Y=Δ^{n_1}×⋯×Δ^{n_k} is a product of k simplices, of total dimension N=n_1+⋯+n_k. We identify the vertices of Y with k-tuples of integers (y_l) such that 0≤y_l≤n_l. We compute Hom(Y,F) as the equalizer of two maps ∏_i F(Δ^N) → ∏_j F(Δ^{N−1}). Here i runs over all nondegenerate N-simplices of Y, which can be identified with (N+1)-tuples of vertices of Y (i.e., k-tuples y as above) such that each k-tuple is obtained from the previous k-tuple by increasing exactly one component by 1. | |
| May 25, 2018 at 21:24 | comment | added | Hadrian Heine | Oh yes. Thanks a lot. So I see. When we have given a finite product of arbitrary representable simplicial sets $Y$ can we also find a functor $H: B \to sSet$ with colimit Y such for every generating cofibration $ i:A→A′$ of $C$ and every map $ i \to Hom(Y,F) $ there is a common lift of all squares $i \to Hom(Y,F) \to Hom(H(b), F) $ for all $b \in B$ that promotes to a map $A' \to \dom( Hom(Y,F))? | |
| May 23, 2018 at 5:01 | comment | added | Dmitri Pavlov | @HadrianHeine: Preservation of Δ^op-indexed homotopy colimits and filtered homotopy colimits implies preservation of all sifted homotopy colimits. Preservation of filtered homotopy colimits is easy to establish directly. | |
| May 23, 2018 at 4:56 | comment | added | Dmitri Pavlov | @HadrianHeine: Instead of d_0 and d_2 we actually have d_1 and d_1, in particualr, α=β, which answers the first group of questions. (The diagonal edge of a simplicial square is the 1st face of both triangles.) | |
| May 22, 2018 at 15:00 | comment | added | Hadrian Heine | And what about sifted categories that are not $\Delta^{op}$, where it is not clear how to decompose products of corepresentables in a nice way to obtain lifts into the limit? | |
| May 22, 2018 at 14:54 | comment | added | Hadrian Heine | We find a lift $q: A' \to dom(F(2)) $ in the square given by $ \alpha \circ \phi : i \to Hom(Y,F) \to F(2)$ as $F(2)$ is a trivial fibration. Why is $ dom(\alpha)∘q = dom(\beta)∘q $ and why is $q$ also a lift in the square given by $ \beta \circ \phi : i \to Hom(Y,F) \to F(2)$? | |
| May 22, 2018 at 14:53 | comment | added | Hadrian Heine | To make sure that I understand you correctly you write $Y:= \Delta^1 \times \Delta^1$ as the pushout of the diagram of maps $d_0 : \Delta^1 \to \Delta^2 $ and $d_2 : \Delta^1 \to \Delta^2$ so that Hom(Y,F) gets a pullback of the diagram $\alpha: F(2)→F(1), \beta: F(2)→F(1). $ Given a generating cofibration $ i : A \to A' $ and a square $\phi: i \to Hom(Y,F) $ we want to find a lift $A' \to dom(Hom(Y,F)) $ in this square. | |
| May 21, 2018 at 22:24 | comment | added | Dmitri Pavlov | @HadrianHeine: Here F(2) and F(1) are some acyclic fibrations in C, and α:F(2)→F(1) is induced by the 1st face map. We know that the pullback F(2) ×_{F(1)} F(2) of F(2)→F(1)←F(2) is also a fibration in C. We must show that it is an acyclic fibration. We verify the lifting property with respect to a generating cofibration i:A→A'. We lift i against F(2), obtaining a map q:A'→dom(F(2)). The map A'→dom(F(2) ×_{F(1)} F(2))=dom(F(2)) ×_{dom(F(1))} dom(F(1)) with components q:A'→dom(F(2)), dom(α)∘q:A'→dom(F(1)), and q:A'→dom(F(2)) is the desired lifting. | |
| May 21, 2018 at 22:16 | comment | added | Dmitri Pavlov | @HadrianHeine: I illustrate it on the easiest nontrivial case when Y=Δ^1×Δ^1, the more general products can be treated analogously. We have Hom(Y,F) = Hom(Δ^2,F) ×_{Hom(Δ^1,F)} Hom(Δ^2,F), which corresponds to the triangulation of a simplicial square using two triangles with a common diagonal. Thus we have a pullback diagram whose legs are F(2)→F(1)←F(2), where both maps are induced by the 1st face map Δ^1→Δ^2. | |
| May 21, 2018 at 22:10 | comment | added | Dmitri Pavlov | @HadrianHeine: Using the Kan recognition theorem for model structures, it suffices to show that if f is a morphism of diagrams such that Hom(Y,f) is a fibration for any finite product of corepresentables Y and Hom(R,f) is a weak equivalence for any corepresentable R, then Hom(Y,f) is an acyclic fibration for any Y. | |
| May 21, 2018 at 14:59 | comment | added | Hadrian Heine | Of course one has a model structure on $C^J$ with generating (trivial) cofibrations the sets of morphisms of the form $ Y×f $, where $f$ is a generating (trivial) cofibration of $C$ and $Y$ is a finite product of corepresentables and where a morphism is a weak equivalence if $Hom(Y,f)$ is a weak equivalence in $C$ for all finite products $Y$ of corepresentables. This model structure has by definition the same cofibrations as the new model structure.To see that both model structures coincide one needs to see that the fibrations or the trivial cofibrations coincide. How can one show this? | |
| May 21, 2018 at 14:45 | comment | added | Hadrian Heine | So by proving that the diagonal functor preserves fibrations one needs at least that a morphism f in $ C^J $ is a fibration in the new model structure if $ Hom(Y,f) $ is a fibration in $C$ for all finite products $ Y $ of corepresentables. How does one get this description of the fibrations in this new model structure? | |
| May 21, 2018 at 0:05 | comment | added | Dmitri Pavlov | @HadrianHeine: This claim is needed to show that the colimit functor preserves acyclic cofibrations. It's easier to show directly that the right adjoint (the constant diagram functor) preserves fibrations. Indeed, fibrations in the new model structure are morphisms f such that Hom(Y,f) is a fibration in C, where Y is a finite product of (co)representables. The constant diagram sends a fibration j to a constant diagram const(j) of such fibrations. Hom(Y,const(j))=Hom(colim(Y),j)=j, as desired. | |
| May 20, 2018 at 22:40 | comment | added | Hadrian Heine | How can one see that a set of generating trivial cofibrations of this model structure is given by the morphisms of the form $ Y \times f, $ where $f $ is a generating trivial cofibration of $C$ and $Y $ is a finite product of corepresentables? | |
| May 20, 2018 at 22:39 | comment | added | Hadrian Heine | To define this modified projective model structure one defines the weak equivalences to be the levelwise ones and the cofibrations to be generated by the morphisms of the form $ Y \times f, $ where $f $ is a generating cofibration of $C$ and $Y $ is a finite product of corepresentables. Then one proves the existence of this model structure by Jardine's theorem. | |
| May 20, 2018 at 22:31 | comment | added | Hadrian Heine | To deduce that the colimit functor $C^J \to C$ is a left Quillen functor, one uses that the generating (trivial) cofibrations of the modified projective model structure can be choosen to be of the form $ Y \times f, $ where $f $ is a generating (trivial) cofibration of $C$ and $Y $ is a finite product of corepresentables. | |
| May 17, 2018 at 5:35 | comment | added | Dmitri Pavlov | @HadrianHeine: Yes, weak equivalences and acyclic fibrations determine a model structure, so it suffices to show that these classes of maps coincide for both model structures. For O-algebras these classes of maps are transferred from the underlying model structure. For J-diagrams with the modified projective model structure weak equivalences are objectwise, whereas acyclic fibrations are natural transformations η such that for any finite product R of representables the map Hom(R,η) is an acyclic fibration. Hom(R,−) takes some limit of maps η(−), and the forgetful functor O-Alg→C creates them. | |
| May 16, 2018 at 19:15 | comment | added | Hadrian Heine | Another remark to the intermediate model structure on J-diagrams in a combinatorial symmetric monoidal model category $C$ you constructed: Given an operad $\mathcal{O} $ in $C$ let's endow the category of J-diagrams in $\mathcal{O}$-algebras with this intermediate model structure. Is it easy to see that this intermediate model structure on J-diagrams in $\mathcal{O}$-algebras is right induced from the intermediate model structure on J-diagrams in $C$? | |
| May 16, 2018 at 5:08 | comment | added | Dmitri Pavlov | @HadrianHeine: One way to proceed is to establish an analog of Proposition 5.7 in the cited paper for restricted Lie algebras, the rest follows formally. Apart from the proof given there, a different (longer, but more direct) proof can be found in the cited work of Harper. Proposition 5.7 is what enables you to prove the condition cited in the first paragraph of my answer. | |
| May 16, 2018 at 4:04 | comment | added | Hadrian Heine | Do you know about a similar statement, where one replaces an admissible operad by a Lawvere theory. I am interested in the question if the forgetful functor from simplicial restricted Lie algebras over a field K of positive char. to simplicial K-vector spaces preserves homotopy geometric realizations. Here the model structures are right induced from the model structure on simplical sets and exist by a theorem of Quillen about model structures on simplicial objects. As far as I know restricted Lie-algebras are not algebras over some operad but only algebras over some Lawvere theory. | |
| May 16, 2018 at 4:03 | comment | added | Hadrian Heine | That looks great. So this new model structure on J-diagrams is monoidal directly from the definition, lies between the projective and injective model structure and exists by the small object argument and Jardine's intermediate model structure theorem. | |
| May 15, 2018 at 20:25 | comment | added | Dmitri Pavlov | (Another point of view on the same model structure: take the closure of simplices under finite products in the category of simplicial sets, and consider diagrams indexed by this category.) | |
| May 15, 2018 at 19:55 | comment | added | Dmitri Pavlov | Indeed, the colimit functor sends these new generating (acyclic) cofibrations to (acyclic) cofibrations: sifted colimits commute with finite products, and the colimit of a representable diagram can be computed by evaluating on the representing object, which yields some generating (acyclic) cofibration in C. | |
| May 15, 2018 at 19:53 | comment | added | Dmitri Pavlov | @HadrianHeine: The above implicitly assumed that J has finite coproducts. However, if this is not the case (such as for J=Δ^op), one can remedy the situation as follows: introduce a new model structure on J-diagrams whose generating (acyclic) cofibrations are defined in the same way as for the projective structure, but with representable functors replaced by finite products of representable functors. By Jardine's intermediate model structure therem this model structure exists. It is monoidal by construction. The colimit functor C^J→C is still a left Quillen functor. | |
| May 14, 2018 at 22:55 | comment | added | Hadrian Heine | As $ C $ is a symmetric monoidal model category, we are reduced to show that for every (trivial) cofibration h of $C$ the map $ (Hom_J(i,-) \times Hom_J(j,-)) \times h $ is a projective (trivial) cofibration. Surely $(Hom_J(i,-) \times Hom_J(j,-)) \times h $ is levelwise a cofibration being the coproduct of cofibrations. Also if $J$ admits finite coproducts, $(Hom_J(i,-) \times Hom_J(j,-)) \times h = Hom_J(i \coprod j,-) \times h $ is a projective cofibration. But how does one conclude in the general case? | |
| May 14, 2018 at 22:54 | comment | added | Dmitri Pavlov | @HadrianHeine: Yes, that's exactly right. | |
| May 14, 2018 at 22:51 | comment | added | Hadrian Heine | Thanks a lot! To make sure that I understand you right let $J$ be a small category and $C$ a cofibrantly generated symmetric monoidal model category. For every $ j \in J$ denote $F_j: C \to C^J$ the left adjoint of evaluation at $ j$. A set of generating (trivial) cofibrations for the projective model structure on $C^J$ is given by $F_j(f)$ for some generating (trivial) cofibration of $C$. For every $c \in C$ we have $F_j (c) = Hom_J(j,-) \times c.$ For all morphisms f, g in $C$ and $i,j \in J$ we have $F_i(f) \square F_j(g) = (Hom_J(i,-) \times Hom_J(j,-)) \times ( f \square g). $ | |
| May 14, 2018 at 21:55 | comment | added | Dmitri Pavlov | @HadrianHeine: Yes, diagrams in a symmetric monoidal model category themselves form a symmetric monoidal model category. The easiest way to see this is to verify that the monoidal product is a Quillen bifunctor by checking the relevant condition on generating cofibrations and generating acyclic cofibrations. These are obtained by tensoring a representable diagram with a generating (acyclic) cofibration in the original category. The pushout product can then be computed as the product of two representable functors tensored with the pushout product of generating (acyclic) cofibrations. | |
| May 14, 2018 at 18:13 | comment | added | Hadrian Heine | Thanks a lot! That sounds great. To deduce that the forgetful functor from diagrams in algebras to diagrams in the underlying symmetric monoidal model category preserves projectively-cofibrant objects, do we need that the projective model structure on diagrams is a symmetric monoidal model category? Can we achieve that the projective model structure on diagrams is symmetric monoidal in this situation? | |
| May 14, 2018 at 5:48 | comment | added | Dmitri Pavlov | @HadrianHeine: I see, this precise question is addressed in Proposition 7.8 in the cited arXiv paper. (See also Theorem 7.10, where it is applied to obtain a comparison result.) In particular, the answer to your question for characterstic 0 chain complexes is positive. | |
| May 13, 2018 at 23:26 | comment | added | Hadrian Heine | I am especially interested in the case of a monad arising from an admissible operad $\mathcal{O} $ in a symmetric monoidal combinatorial model category, where all objects are cofibrant. In the easiest example let $\mathcal{C} $ be the category of chain complexes over a field of char 0 and $\mathcal{O} $ the Lie-operad. | |
| May 13, 2018 at 20:36 | comment | added | Dmitri Pavlov | @HadrianHeine: I misstated the condition, it's actually a bit more complicated. I adjusted my writeup. Do you have a specific monad T in mind? | |
| May 13, 2018 at 20:36 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| May 13, 2018 at 16:17 | comment | added | Hadrian Heine | Let's assume that T preserves cofibrant objects and let J be a small sifted category and $H: J \to Alg_T(\mathcal{C}) $ a functor that is projectively-cofibrant in $Alg_T(\mathcal{C})^J.$ Denote $V: Alg_T(\mathcal{C}) \to \mathcal{C} $ the forgetful functor. Why is $ V(hocolim H) \simeq V(colim H) \simeq colim(VH) $ the homotopy colimit of VH? Is VH projectively cofibrant in $\mathcal{C}^J$? If yes, why? Does the forgetful functor preserve projectively cofibrant diagrams if T preserves cofibrant objects? If yes, how does one show this? | |
| May 13, 2018 at 15:58 | comment | added | Hadrian Heine | That sounds very interesting to me. | |
| May 12, 2018 at 17:11 | history | answered | Dmitri Pavlov | CC BY-SA 4.0 |