Recall the acyclic models theorem: given two functors $F, G$ from a "category $\mathcal{C}$ with models $M$" to the category of chain complexes of modules over a ring $R$, a natural transformation $H_0(F) \to H_0(G)$ induces a natural transformation $F \to G$ (unique up to natural chain homotopy) if $F$ is free on the models and $G$ is acyclic. (This has many useful applications, such as showing that things like the Alexander-Whitney maps exist and are unique up to homotopy.)

The proof is a bit of homological algebra, but I suspect that there may be a fancier homotopical way to think about it. Namely, I want to say something of the form that the category of functors from $\mathcal{C}$ to chain complexes has a model structure on it, the free functors are cofibrant in some sense, and the map $G \to H_0(G)$ should be an acyclic fibration. Thus the lifting of $F \to H_0(F) \to H_0(G)$ to $F \to G$ would just be a lifting argument in a general model category (as would be the existence of a natural chain homotopy). However, I'm not sure what the model structure should be. The projective model structure doesn't seem to be the right one because $G \to H_0(G)$ is not a weak equivalence of chain complexes (it only is for the models $M$). So I guess the model structure relevant here would be a hybrid of the projective model structure, where instead of weak equivalences and fibrations levelwise, one would just want them to hold on $M$: that is, a morphism $A \to B$ is a weak equivalence (resp. fibration) if and only if $A(m) \to B(m)$ is one for each $m \in M$. It seems that the lifting property needed is essentially the proof of the acyclic models theorem itself.

I strongly suspect this can be done. Am I right? Can this be pushed further?

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    $\begingroup$ +1, I think this is a great question. I'm going to go home and think about it tonight! $\endgroup$ Sep 23, 2011 at 0:48
  • $\begingroup$ Just a guess: are Reedy model structures helpful? ncatlab.org/nlab/show/Reedy+model+structure $\endgroup$
    – DamienC
    Sep 23, 2011 at 8:47
  • $\begingroup$ I'm not really sure how. At least for the applications I have in mind (e.g. constructing the product maps or the Steenrod operations), $\mathcal{C}$ should be something like the category of topological spaces or simplicial sets (or simplicial abelian groups), which is not a Reedy category. $\endgroup$ Sep 23, 2011 at 13:18
  • $\begingroup$ Just for completeness, since the Wikipedia article linked above leaves a lot to be desired, here is a more detailed source on the Acyclic Models Theorem and on Categories with Models: amathew.wordpress.com/2010/09/11/the-method-of-acyclic-models $\endgroup$ Jul 24, 2012 at 21:28
  • $\begingroup$ A very late comment: I would like to add another possible viewpoint of the acyclic models theorem: by Dold-Kan correspondence, instead of connective chain complexes, we can consider the simplicial abelian groups. It seems to me that Proposition in Lurie's Higher Algebra version Sept 2017 could be slightly modified to be a generalization of the acyclic models theorem. $\endgroup$
    – user20948
    Nov 28, 2018 at 12:30

2 Answers 2


Akhil's answer works proposes a model structure, but it doesn't have the property required by the original question, i.e. that free functors are cofibrant. Also, proving factorization is much more involved than Akhil's answer would lead you to believe (in fact, this is where all the hard work is done). In this answer I'll fill in the details of that proof and at the bottom address the non-cofibrancy of free functors.

The proof that the model structure exists mimics Hirschhorn's proof that the projective model structure exists, and that proof takes about 6 pages. I'll summarize it here, but those who know it can just skip the next section where I fill in the details in Akhil's setting (i.e. with fibrations and weak equivalences defined based on $M\subset K$). The last section brings it back to the acyclic models theorem.

Projective Model Structure

The reason factorization works in the standard projective model structure is that you transfer the model structure from the cofibrantly generated $C^{K^{disc}} = \prod_{ob(K)} C$ along a Quillen adjunction $(F,U)$, and you apply the small object argument. This transfer principle is Hirschhorn's Theorem 11.3.2. Let $I$ and $J$ be the generating cofibrations and generating trivial cofibrations for $C$. For every $k\in K$ let $I_k= I \times \prod_{k'\neq k} id_{\phi}$, where $\phi$ is the initial object of $C$. This is how to view $I$ in just one component of $C^{K^{disc}}$. The generators for $C^{K^{disc}}$ are $I_{ob(K)} = \cup_{k\in K} I_k$ and similarly $J_{ob(K)} = \cup_{k\in K} J_k$.

The existence of the projective model structure (and the fact that it's cofibrantly generated) is Hirschhorn's Theorem 11.6.1 and the proof verifies that the hypotheses of 11.3.2 hold, namely:

  1. $F(I_{ob(K)})$ and $F(J_{ob(K)})$ permit the small object argument.
  2. All maps in $U(F(J_{ob(K)})-cell)$ are weak equivalences.

For Hirschhorn, the functor $F$ takes a discrete diagram $X$ to a diagram $\coprod_{k \in K} F^k_{X(k)}$ in $C^K$ where $F^k_A = A \otimes K(k,-)$. The functor $U$ is the forgetful functor. Hirschhorn's proof of (2) requires Lemma 11.5.32, which analyzes pushouts of $F_A^k \to F_B^k$, since for any $g:A\to B$, $F(g\otimes \prod_{k'\neq k} id_\phi)$ is the map $F_A^k\to F_B^k$. The point of Lemma 11.5.32 is that if $X\to Y$ is a pushout of this map then $X(k)\to Y(k)$ is a pushouts in $C$ of a giant coproduct of copies of $g$. This coproduct is a trivial cofibration, so pushouts of it are trivial cofibrations and hence $X\to Y$ is a weak equivalence as desired.

Lemma 11.5.32 gets used again (along with the Yoneda lemma and basic facts about smallness) to prove (1). The point is that the domains of the generators are $F_A^k$ and $A$ is small relative to $I$-cof, so one can use adjointness to pass between maps in $C^K$ and maps in $C$ (i.e. one component of $C^{K^{disc}}$) and use smallness there to get the desired commutativity of colimit and hom.

Modified Projective Model Structure

To prove Akhil's proposed model structure has factorization you need to define the correct functor $F$ AND you have to verify (1) and (2). It turns out that Akhil has the correct $F$ (if you soup it up so it goes from $C^{K^{disc}}\to C^K$ rather than $C\to C^K$), but verifying (1) and (2) are non-trivial because this $F$ doesn't automatically have Lemma 11.5.32. Anyway, the correct $F$ takes $X$ to $\coprod_{m\in M}F^m_{X(m)}$ in $C^K$. Note that it doesn't land in $C^M$ because $K(m,-)\neq M(m,-)$. This is good, since if it landed in $C^M$ you'd just end up with the projective model structure on $C^M$.

To make a long story short, you can prove an analog of Lemma 11.5.32 and follow Hirschhorn to get the modified projective model structure. A better way is to notice that on the coordinates in $K - M$ all maps are trivial fibrations, the cofibrations are isomorphisms, and both the generating cofibrations and the generating trivial cofibrations are $\lbrace id_\phi \rbrace$. So if $M$ is empty you get the trivial model structure (Hovey's example 1.1.5) on $C^K$, and if $M$ is a one-point set you get a product of $C$ with a bunch of copies of that trivial model structure. Of course if $M=K$ you get the usual projective model structure.

With this observation, a better way to verify existence of this modified projective model structure is to take a product where in coordinate $k\in M$ one uses the projective model structure and in coordinate $k\not \in M$ one uses the trivial model structure. Note that this modified projective structure is done in Johnson and Yau's "On homotopy invariance for algebras over colored PROPs" for the case where $K$ is a group acting on $C$ and $M$ is a subgroup. Unfortunately, I only found this reference after working the above out for myself.

Connection to Acyclic Models Theorem

The main reason why I don't think this model structure helps with the acyclic models theorem is what I alluded to in my comment to Akhil's answer. You really need $K$ to be small, and this rules out all the interesting examples to which the acyclic models theorem applies. The other reason is the observation above about getting the trivial model structure in coordinates $k\notin M$. With this, you can't have the "free functors being cofibrant" that the OP asks for. Recall that a free functor here means there are models $M_\alpha \in M$ and $m_\alpha \in F(M_\alpha)$ such that for all $X\in K$ the set $\lbrace F(f)(m_\alpha)\rbrace$ is a basis for $F(X)$. Because cofibrations are isomorphisms in the coordinates $k\notin M$, the only cofibrant object there is the initial object, i.e. the empty functor. These free functors are not of that form because their behavior is not restricted in that way on $k\notin M$.

  • $\begingroup$ In figuring out how to do the modified projective model structure I also tried to skip over having to mimic Hirschhorn's proof by way of undercategories. I had hoped the resulting model structure would be either $C^{M/K}$ or $C^M/C^K$. Neither works. The first disallows diagrams with nontrivial entries in the $k$-th spot unless $k$ is the codomain of some morphism with domain in $M$. The second isn't a model category unless $M$ is a one-point set (in which case you get $C^K$ again). I couldn't find anywhere undercategories were studied under a set of objects instead of just one. Anyone know? $\endgroup$ Aug 10, 2012 at 17:36
  • $\begingroup$ Hi David. Thanks for your comments and this detailed answer. It's been a little while since I've thought about this. I'm rather tired at the moment; I'll go through this more carefully tomorrow. $\endgroup$ Aug 11, 2012 at 2:41
  • $\begingroup$ I just want to point out that there is another well-oiled machine one can use to get the existence of the model structure --- Prop A.2.6.8 in "Higher Topos Theory." It says roughly that when you have a reasonable class of cofibrations and a reasonable class of weak equivalences in a presentable category, then you get a combinatorial model structure once you know that any trivial fibration is a weak equivalence. "Reasonable" here means that one has to check certain set-theoretic conditions (e.g., that the cofibrations are generated by a set). If you look at Lemma A.2.8.3... $\endgroup$ Aug 11, 2012 at 12:55
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    $\begingroup$ Anyway, either way, I think the model structure we want definitely does exist. Here's a question that I just thought of: what is the associated $\infty$-category? The projective (or injective) model structure is a presentation for the $\infty$-category of presheaves (of spaces) on a category; what does this model structure present? I don't know, but I'll think about it. $\endgroup$ Aug 11, 2012 at 13:18
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    $\begingroup$ I learned recently that this model structure is a special case of one due to Mark Johnson: On modified Reedy and modified Projective model structures, Theory and Applications of Categories, 24 (2010) 179-208. $\endgroup$ Jan 17, 2018 at 14:20

Here's the model structure that I think works. Let $\mathcal{C}$ be a cofibrantly generated model category, with generating cofibrations $I$ and generating trivial cofibrations $J$. Let $K$ be a small* category with models $M$ (so $M$ is a set of objects in $K$). We want to construct a cofibrantly generated model structure on $\mathcal{C}^K$, the category of functors $K \to \mathcal{C}$, such that a weak equivalence (resp. fibration) of functors $F \to G$ is one such that $Fm \to Gm$ is one such for $m \in M$.

Namely, let's define fibrations and weak equivalences in that way. Define cofibrations (or trivial cofibrations) by use of the appropriate lifting property. The claim is that if $m \in M$, and $x \to y$ is in $I$ (resp. $J$), then $\hom(m, \cdot) \times x \to \hom(m, \cdot) \times y$ is a cofibration (resp. trivial cofibration) in $\mathcal{C}^K$. This follows because lifting with respect to a diagram $F \to G$ is the same as finding a lifting in the diagram in $\mathcal{C}$ with $x \to y$ versus $Fm \to Gm$ (by Yoneda's lemma). Moreover, we see that these generate the cofibrations (resp. trivial cofibrations) in $\mathcal{C}^K$ in the sense that anything having the right lifting property with respect to these morphisms is necessarily one such that $Fm \to Gm$ is a trivial fibration (resp. fibration) in $\mathcal{C}$ for each $m$, so is a trivial fibration (resp. fibration) in the functor category.

So the lifting axiom is clear. The retract axiom is easy to verify, and the factorization axiom follows from the small object argument applied to the maps $\hom(m, \cdot) \times x \to \hom(m, \cdot) \times y$. This shows that one actually has a model structure.

In short, one can just imitate the projective model structure by loosening the notions of fibration and weak equivalence. When one takes $\mathcal{C}$ to be the category of chain complexes with the usual model structure, then the acyclic model theorem is just the lifting axiom applied to the model structure on $\mathcal{C}^K$, as described in the question.

*There are set-theoretic issues here when $K$ is something like spaces or simplicial sets, of course. I'm not really well-versed enough in the details to know what to do that works reasonably other than appeal to universes (although this is not necessary).

  • $\begingroup$ I don't think you're going to have any luck at all if $K$ is not small. I mean, $C^K$ isn't even a category in that case because you have more than a class worth of objects. However, you can do slightly better (e.g. small presheaves over large categories where $K$ is only locally small) using the framework of Chorny and Rosicky in arxiv.org/abs/1110.0605 and arxiv.org/abs/1110.4252. They develop a new notion of locally presentable which works in that setting, then get a notion of what it means to be combinatorial, then do homotopy with it in the spirit of Jeff Smith. $\endgroup$ Jul 24, 2012 at 20:54
  • $\begingroup$ Thanks again for asking such an interesting question. I had fun thinking about it. By the way, there is a small typo in your answer. In the second paragraph, $hom(m,\dot,y)$ should be $hom(m,\dot)\times y$. Also, "resp cofibration" should be "resp trivial cofibration" $\endgroup$ Aug 10, 2012 at 17:32

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