Do homotopy pullbacks always compose?

Question

Classical pullbacks compose, as is easily checked with the universal property. More precisely, if $\require{AMScd} \begin{CD} A @>>> B\\ @VVV @VVV\\ C @>>> D \end{CD}$ and $\require{AMScd} \begin{CD} B @>>> E\\ @VVV @VVV\\ D @>>> F \end{CD}$ are pullback diagrams, then so is $\require{AMScd} \begin{CD} A @>>> E\\ @VVV @VVV\\ C @>>> F \end{CD}$.

I was wondering whether this was true for homotopy pullbacks, and if so, with what level of generality.

For instance if you take the usual model for homotopy pullbacks in $\mathbf{Top}$, you can get a very concrete homotopy equivalence between the two homotopy pullbacks.

Also if you have a model structure in which, to compute homotopy pullbacks it suffices to replace one map by a fibration and then take the usual pullback, then you can show by checking universal properties that this still works (I was told that perhaps something like "if you have a Reedy model structure on $C^I$ then it works" would work -but I don't know what that means yet)

So the most general setting I can think of, to make sense of the question is : we have a homotopical category $(C,W)$ (that is, a category $C$ with a wide subcategory of weak equivalences, satisfying the 2-out-of-3 or 2-out-of-6 property) and our diagram category $I=\require{AMScd} \begin{CD} &&\bullet\\ & @VVV\\ \bullet @>>> \bullet \end{CD}$ and then $C^I$ is also a homotopical category with pointwise weak equivalences; and we assume $\lim : C^I\to C$ has a right derived functor $\mathbb{R}\lim : C^I\to C$ (with Riehl's terminology in Categorical homotopy theory), then we call $\mathbb{R}\lim (\require{AMScd} \begin{CD} && B\\ & @VVV\\ C @>>> D \end{CD})$ a homotopy pullback of $B,C$ over $D$.

It seems that with this level of generality, I can't get a map from the homotopy pullback to its components (or at least I don't see how) : I get a map in $\mathrm{Ho}(C)$ by the universal property of the Kan extension and by seeing the components as homotopical functors $C^I\to C$, but this map is a priori only a zigzag of maps in $C$; so first of all for the question to make sense we have to figure out a setting in which we do get honest maps from the homotopy pullback to its components, that is, maps in $C$.

I don't really see how to get that so :

What are some natural conditions we can impose on the situation to get natural maps in $C$ from the homotopy pullback to the components of the diagram ? Natural lifts of the natural transformation $\delta\mathbb{R}\lim \implies \delta\pi$ where $\delta : C\to \mathrm{Ho}(C)$ is the localisation functor, and $\pi$ is any of the "component of the diagram" functors ?

Once we have these conditions we can phrase our question :

do homotopy pullbacks compose ? More precisely, if we have two homotopy pullback diagrams as in the very beginning of this question, with the maps $B\to E, D$ and $A\to B,C$ the natural maps induced by the conditions; when is there an isomorphism in $\mathrm{Ho}(C)$ from $A$ to the homotopy pullback of $E,C$ over $F$ making the obvious diagram commute ? When is this isomorphism (or its inverse) a map in $C$ ?

EDIT : See Pece's answer for another formulation of the question: I will also accept answers that answer that other formulation.

Answer 1

This is not really an answer but a long remark on your setting. It seems quite unnatural to look for genuine maps in $C$ from the homotopy pullback to its component. Your question can be stated without this requirement.

For a category $C$ with a class of weak equivalences $W$ (we can drop any requirement on $W$ as long as our set-theoretic foundations allow for the Gabriel-Zisman localization $W^{-1}C$ to exist), we can define the homotopy limit functor of shape $I$ as follows: the category "constant diagram" functor $\delta : C \to C^I$ respects weak equivalences if we take the pointwise weak equiavlences in $C^I$, so that it induces a functor $$ \boldsymbol \delta : W^{-1}C \to {(W^I)}^{-1}(C^I)$$ The homotopy limit functor is the right adjoint to this functor (when it exists).

Suppose all homotopy pullbacks existin $C$ and take a commutative diagram in $C$: $$\require{AMScd} \begin{CD} A @>>> B\\ @VVV @VVV\\ C @>>> D \end{CD}$$ Denote $H$ for its homotopy pullback. The commutative diagram is in particular a map $\boldsymbol \delta A \to \begin{CD} {} && B\\ {} @VVV\\ C @>>> D \end{CD} $ in ${(W^I)}^{-1}(C^I)$ so it corresponds a unique map $A \to H$ in $W^{-1}C$. You can say that the original square in $C$ is homotopy cartesian if the map $A \to H$ is an isomorphism. Then it makes sense to ask if the composition of two homotopy cartesian square is again homotopy cartesian.

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Now I don't have the answer to your question in full generality, but it seems to be true in the case of model categories (based on a few scribbling so don't take my word for it), and the key seems to be that the map $A\to H$ is represented in $C$ (by which I mean that it is the image of a genuine $A \to H$ in $C$ through the localization functor).

Answer 2

Expanding on my comment under another answer, suppose your relative category makes the 2-functor $I\mapsto C^{I}[(W^I)^{-1}]$, where $I$ is any finite direct (aka homotopically finite) category, into a right derivator. This should be interpreted as saying that $C$ admits homotopy right Kan extensions along all functors between finite direct categories, in particular, $C$ admits homotopy pullbacks. Then a standard result, which can be found as 3.14 in Groth's paper linked above, says that cartesian squares paste. The cartesian squares in this case are determined by exactly the property in Pece's answer, that the induced map to the summit of the homotopy pullback square on the same span be an isomorphism in $C[W^{-1}]$. That $(C,W)$ induces a derivator is not easy to prove, however. I'm not I can think of any examples except when $(C,W)$ is essentially a category of fibrant objects.