Deriving the functor $ \int_{\Gamma} F(-,-)$

Question

Suppose that $C$, $D$, and $E$ are combinatorial model categories, so that for any category $\Gamma$, the functor categories $C^{\Gamma}$, $D^{\Gamma}$, and $E^{\Gamma}$ have both the projective and injective model structures. Suppose that furthermore $F\colon C^{\mathrm{op}}\times D\to E $ is any bifunctor having a (point-set) right derived functor (in the sense of e.g. Shulman, definition 2.5) of the form $H(Q(-), R(-))$ for some $H\colon C^{\mathrm{op}}\times D\to E$ and cofibrant resp. fibrant replacement functors $Q, R$.

Now consider the end construction $\int_{\Gamma} F(-,-)\colon (C^{\Gamma})^{\mathrm{op}}\times D^{\Gamma} \to E$, and endow $C^{\Gamma}$ and $D^{\Gamma}$ with the projective model structures.

Question: Will $\int_{\Gamma} H(Q_{\mathrm{proj}}(-),R_{\mathrm{proj}}(-))\colon (C^{\Gamma})^{\mathrm{op}}\times D^{\Gamma} \to E $ be a right derived functor of $\int_{\Gamma} F$ (where $Q_{\mathrm{proj}}$ and $R_{\mathrm{proj}}$ are cofibrant resp. fibrant replacements in the projective model structure)?

In the case where $F$ is part of a two-variable Quillen adjunction, this already appears to be the case, c.f. nLab – Quillen bifunctor (one has to dualize). I am interested in whether it still holds without this assumption. At the end of the day, I am interested in the case $ F = \mathbf{dgFun}\colon\mathbf{dgCat}^{\mathrm{op}}\times\mathbf{dgCat}\to\mathbf{dgCat}$ and $ H = \mathbf{dgFun_{\infty}}$, where $\mathbf{dgFun_{\infty}}(A,B)$ has objects dg-functors $ A \to B $ and morphisms $A_{\infty}$-natural transformations between such. Here we regard $\mathbf{dgCat}$ as a 1-category endowed with Tabuada’s model structure.

Note: As the above question probably indicates, I am no expert in this field, I am learning it in fact, so please feel free to point out anything that might be wrong with my question.

EDIT: The answer in the case $F = H = \mathbf{dgFun_{\infty}}$ would also be sufficient.

Answer 1

We eventually found a solution which solved the original issue, though not in the way it is exactly stated here. Namely, we found what would appear to be the most general way that ends can be derived. Specifically,

Theorem: Let $\mathscr{C}$ be a model category and $\Gamma$ a category. Regard the functor category $\mathscr{C}^{\Gamma^{\mathrm{op}}\times\Gamma}$ as a model category in any of the following ways:

1. as $\mathscr{C}^{\Gamma^{\mathrm{op}}\times\Gamma} =(\mathscr{C}^{\Gamma}_{\mathrm{Proj}})^{\Gamma^{\mathrm{op}}}_{\mathrm{Inj}}$ (assuming this model structure exists);

2. as $\mathscr{C}^{\Gamma^{\mathrm{op}}\times\Gamma} =(\mathscr{C}^{\Gamma^{\mathrm{op}}}_{\mathrm{Proj}})^{\Gamma}_{\mathrm{Inj}}$ (assuming this model structure exists);

3. As $\mathscr{C}^{\Gamma^{\mathrm{op}}\times\Gamma} =\mathscr{C}^{\Gamma^{\mathrm{op}}\times\Gamma}_{\mathrm{Reedy}}$ (assuming $\Gamma$ is a Reedy category).

Then the end functor $\int_{\Gamma}\colon \mathscr{C}^{\Gamma^{\mathrm{op}}\times\Gamma}\to \mathscr{C}$ is right Qullen.

Despite its simple proof, we were surprised not to find this result written anywhere, and several experts on homotopy theory claimed they had not seen it anywhere either. We therefore wrote a short paper on it, see https://arxiv.org/abs/1807.03266. It is very close to known results, though, and we are more than ready to accept that this result is already known and written somewhere and that we simply could not find it. If this is the case, we shall be happy to receive a reference.

Answer 2

A long comment.

1. The way you state the question currently mixes the abstract notion of derived functor (has nothing to do with fibrant replacements), and the notion of a co/fibrant replacement. I am not sure how to relate the two without some additional condition. However, given $X \in \cal M$, its category of, say, cofibrant replacements $\{ QX \stackrel \sim \to X \}$ is contractible (2.4.8 in this paper, there is also a proof of Hinich of the same fact). This is what in practice guarantees that for $F: \cal M \to \cal D$ sending weak equivalences of cofibrants to isomorphisms, the assignment $X \mapsto F(QX)$ is independent of choices and gives the left derived functor of $F$: the verification of the universal property is almost immediate. Of course, the formula you write usually works in practice, but the ways things are proven is more like this:

2. A statement that is model-categorical in flavour is (the dual of) A.2.9.26 in Higher Topos Theory; the definition of Lurie is minimal in character, he just requires the preservation of colimits, not adjoints. Perhaps it is possible to get away with less than all colimits, especially if the Reedy category in question is simple enough. I have not checked.

3. For a flavour of homotopy co/end theory without replacements and model structures, one could use derivators, as in section 5 of this paper, though they are not exactly answering your question, either.

Answer 3

The answer is no in this generality, but I do not know what happens specifically for ${\bf dgCat}$ and ${\bf dgFun}$. For convenience, let me construct a counter-example to the dual of your question, where $F$ is covariant in both $C$ and $D$, the end is replaced by a coend, and we use the projective model structure on $D^{\Gamma}$ and the injective model structure on $C^{\Gamma^{\rm op}}$ (you can then obtain a counter-example to your question by replacing $C$ with $C^{\rm op}$).

Let ${\cal M}$ be the combinatorial model category of simplicial groups and let $F: {\cal M} \times {\cal M} \to {\cal M}$ be the functor $F(A,B) = A \times B$. Then $F$ is not a left Quillen bifunctor but it does preserve weak equivalences in each variable separately and is hence its own derived functor. Now let $\Gamma = [1]$ be the category with two objects and one non-identity morphism. Then the coend functor $$\int F(-,-): {\cal M}^{[1]^{\rm op}} \times {\cal M}^{[1]} \to {\cal M}$$ is given by the pushout-product $$ (B \leftarrow A,C \rightarrow D) \mapsto A\times D \coprod_{A \times C} B \times C $$ in the category of simplcial groups. If the answer to the dual version of your question was positive it would imply that the pushout above has the "right type" as soon as $A,B,C,D$ are all cofibrant and $C \to D$ is a cofibration. But this is not true. For example, suppose that $G = \langle g\rangle$ is the free group on a single generator (considered as a discrete simplicial group), and let $f$ and $g$ be the maps $\{1\} \leftarrow G$ and $\{1\} \rightarrow G$ respectively. Then $G$ is cofibrant (since it is free) and so $g$ is a cofibration. The coend of $f$ and $g$ is given by $G\times G \coprod_{G \times \{1\}}\{1\} = G$. Let us now factor $f$ as $\{1\} \stackrel{\simeq}{\leftarrow} H \stackrel{f'}{\leftarrow} G$ where $f'$ is a cofibration of simplicial groups. Then the coend of $f'$ and $g$ is given by the pushout $G\times G \coprod_{G \times \{1\}}H \times \{1\}$, which is also a homotopy pushout since simplicial groups is a left proper model category and $f': G \to H$ is a cofibration. To compute this homotopy pushout one can use the fact that simplicial groups are Quillen equivalent to reduced simplicial sets by means of the $\overline{W}$-construction and Kan loop group adjunction, where $\overline{W}$ is a particular model for the classifying space functor, and the Kan loop group functor $\Omega$ is a particular model for the loop functor. The above homotopy pushout is then equivalent to the Kan loop group of the homotopy pushout of reduced simplicial sets $$ \overline{W}(G\times G) \coprod^h_{\overline{W}G \times \{1\}}\overline{W}H \times \{1\} = [S^1 \times S^1] \coprod^h_{S^1 \times \ast} \ast \simeq S^1 \vee S^2 .$$ The map $\Omega(S^1 \vee S^2) \to G \simeq \Omega(S^1)$ induced by the weak equivalence $[H \leftarrow G] \to [\{1\} \leftarrow G]$ is the map induced by the collapse map $S^1 \vee S^2 \to S^1$, and is clearly not a weak equivalence of simplicial groups. This means that taking coend against the projectively cofibrant object $[\{1\} \to G] \in {\cal M}^{[1]}$ does not preserve weak equivalences between injectively cofibrant objects in ${\cal M}^{[1]^{\rm op}}$, and so the suggested construction is not homotopical in general.