The next step (in our discussion, started yesterday, of the cotangent complex) will be to define a model structure on the category of algebras over a fixed ring. Model structures allow one to define derived functors in a non-abelian setting. The key idea is that, when you want to derive an additive functor $F$ on an object $X$ in some abelian category, you replace $X$ by a projective resolution and evaluate the functor $F$ on this resolution. (And then, take its homology; in the setting of derived categories, though, one usually just takes $F$ of the projective resolution and leaves it at that.) Because $F$ on projective resolutions is much better behaved than $F$ simply on modules, the derived functor is a nice replacement.

The intuition is that a projective resolution is a cofibrant approximation to the initial object, in the language of model categories (which is often seen as a non-abelian version of classical homological algebra). This is actually precisely true if one imposes the usual model structure on bounded-below chain complexes for modules over a ring, for instance.

In constructing the cotangent complex, we are trying to derive the (highly non-abelian) functor of abelianization, which as we saw was closely related to the construction of differentials. This functor was defined on rings under a fixed ring $A$ and over a fixed ring $B$, which is not anywhere near an abelian category. So we will need the language of model categories, and today we shall construct a model structure on a certain class of categories.

In deriving an additive functor, one ultimately applies it not on the initial abelian category, but the larger category of chain complexes. Here the analogy extends again: by the Dold-Kan correspondence (which I recently talked about), this is equivalent to the category of simplicial objects in that category. The appropriate approach now seems to be to define a model structure not on $A$-algebras over $B$, but on the category of simplicial $A$-algebras over $B$.

1. Lifting model structures via adjunctions

We are now interested in obtaining a model structure on the category of simplicial ${A}$-algebras. We shall take as known that there exists a cofibrantly generated model structure on the category ${\mathbf{SSet}}$ of simplicial sets, whose cofibrations are the injections, fibrations are Kan fibrations, and weak equivalences are maps inducing homotopy equivalences on the geometric realizations.

Now we know that there is an adjunction

$\displaystyle \mathbf{S}\mbox{--}\mathbf{Alg}^A \rightleftarrows \mathbf{SSet}$

that consists of the forgetful functor and the free algebra functor. From this adjunction, we want to induce a model structure on ${\mathbf{S}\mbox{--}\mathbf{Alg}^A}$. We will do so using the following statement:

Theorem 5 (Quillen) Suppose ${\mathcal{A}, \mathcal{B}}$ are complete and cocomplete categories all of whose objects are small, and suppose ${F: \mathcal{A} \rightarrow \mathcal{B}, G: \mathcal{B } \rightarrow \mathcal{A}}$ are a pair of adjoint functors.Suppose ${\mathcal{A}}$ is a cofibrantly generated model category with a set ${S}$ of generating cofibrations and a set ${T}$ of generating acyclic cofibrations. Suppose moreover ${GX}$ is a fibrant object in ${\mathcal{A}}$ for any ${X \in \mathcal{B}}$. Then there exists a cofibrantly generated model structure on ${\mathcal{B}}$, generated by ${FS}$, such that

1. The weak equivalences ${w}$ in ${\mathcal{B}}$ are those such that ${Gf}$ is a weak equivalence.
2. The fibrations ${f}$ in ${\mathcal{B}}$ are those such that ${Gf}$ is a fibration.
3. The cofibrations are thus determined (as those having the left lifting property with respect to the acyclic fibrations).

The pair ${(F, G)}$ is then a Quillen adjunction between ${\mathcal{A}, \mathcal{B}}$.

Proof: There are several axioms to check. First, we have assumed ${\mathcal{B}}$ is complete and cocomplete, so the axioms of the additional model structure are left to check.

We start by noting that if ${c: A \rightarrow A'}$ is an acyclic cofibration in ${\mathcal{A}}$, then ${Fc}$ has the llp (left lifting property) with respect to all fibrations in ${\mathcal{B}}$. For indeed, if ${p: X \rightarrow X'}$ is a fibration in ${\mathcal{B}}$, then by definition ${Gp}$ is a fibration in ${\mathcal{A}}$, and the lifting problem

is equivalent to the lifting problem

The latter, however, will admit a solution if ${c}$ is an acyclic cofibration in ${\mathcal{A}}$, because ${\mathcal{A}}$ already satisfies the model category axioms. It follows that ${F}$ sends acyclic cofibrations in ${\mathcal{A}}$ to maps having the llp with respect to all fibrations in ${\mathcal{B}}$. Similarly ${F}$ sends cofibrations in ${\mathcal{A}}$ to cofibrations in ${\mathcal{B}}$ (because ${F}$ will send a cofibration in ${\mathcal{A}}$ to something having the llp for all acyclic fibrations in ${\mathcal{B}}$, which is the definition of a cofibration).

The first claim is that any morphism in ${\mathcal{B}}$ admits a functorial factorization into a cofibration followed by an acyclic fibration. To see this, apply the small object argument to the set of maps ${FS}$. We find that any map ${X \rightarrow Y}$ in ${\mathcal{B}}$ can be factored ${p \circ i}$ such that ${i}$ is a transfinite composition of pushouts of maps in ${FS}$ (hence a cofibration in ${\mathcal{B}}$) and ${p}$ has the left lifting property with respect to ${FS}$.

But this implies that ${p}$ is a fibration; indeed, by the same argument as before, we see that since ${p}$ has the llp with respect to ${FS}$, ${Gp}$ has the llp with respect to ${S}$. Thus ${Gp}$ is an acyclic fibration, and by definition so is ${p}$. So we get one of the (functorial) factorizations.

By the same reasoning, applying the small object argument to ${FT}$, we find that any map ${X \rightarrow Y}$ in ${\mathcal{B}}$ can be factored as ${q \circ j}$ such that ${j}$ is a transfinite composition of maps in ${FT}$ (in particular, a cofibration) and ${q}$ has the llp with respect to ${FT}$; in particular, ${Gq}$ has the llp with respect to ${T}$, and is thus a fibration. We want to say that ${j}$ is an acyclic cofibration, but in general we don’t know that ${F}$ preserves weak equivalences. However, at least we can say that ${j}$ has the llp with respect to all fibrations, because it is a transfinite composition of pushouts of maps of ${FT}$, and every map in ${T}$ has the llp with respect with respect to all fibrations. We will show below that anything with the llp with respect to all fibrations (in ${\mathcal{B}}$) is in fact a weak equivalence. So ${j}$ is, indeed, an acyclic cofibration.

That the two-out-of-three axiom holds for weak equivalences in ${\mathcal{B}}$ is immediate. Finally, we need to see that the lifting axioms hold: given a diagram in ${\mathcal{B}}$

with ${i}$ a cofibration and ${p}$ a fibration, then if one of ${i,p}$ is a weak equivalence, we must show that a lift ${B \rightarrow X}$ exists.

One direction is straightforward. If ${p}$ is a weak equivalence, then the definition of a cofibration in ${\mathcal{B}}$ shows that a lift exists. Now suppose ${i}$ is the acyclic one. We can factor ${i}$ as a composite ${i_1 \circ i_2}$ where ${i_1}$ is a fibration and ${i_2}$ has the llp with respect to all fibrations (by using the above small object argument on ${FT}$ as above so that ${i_2}$ is a transfinite composite of pushouts by ${FT}$). The lemma that we have not proved yet will imply that ${i_2}$ is a weak equivalence.

Now ${i}$ will be a retract of ${i_2}$. Indeed, this is a standard argument. Let ${i: A \rightarrow B}$ and ${i_2: A \rightarrow C, i_1: C \rightarrow B}$. We have a diagram

By two-out-of-three, ${i_1}$ is an acyclic fibration, so a lift ${B \rightarrow C}$ exists. This means precisely that ${i}$ is a retract of ${i_2}$. But as ${i_2}$ has the llp with respect to all fibrations (in view of its construction), so does ${i}$. This completes the verification of the lifting axiom.

Hence we need:

Lemma 6 Hypotheses as above, any morphism ${i: A \rightarrow B}$ in ${\mathcal{B}}$ with the llp with respect to all fibrations is an acyclic cofibration.

Proof: By assumption it is a cofibration, though we need to see that it is a weak equivalence.

Now by assumption, ${A}$ is fibrant, so the diagram

and the assumption about ${i}$ shows that there is a retraction ${r: B \rightarrow A}$. We thus get two morphisms ${B \rightrightarrows B}$ given by ${1_B}$ and ${ir}$. Ideally, we want to claim that the two maps ${1_B, ir}$ are homotopic. In other words, we want a factorization

Here ${PB}$ is a path object for ${B}$.

In order to do this, we consider the diagram

Here ${A \rightarrow PB}$ is the composite ${A \rightarrow B \stackrel{s}{\rightarrow} PB}$, which makes sense there is always a natural map ${B \rightarrow PB}$ (which by definition is required to be a weak equivalence). Since ${iri = i}$, the diagram commutes.

Since ${i}$ has the llp with respect to all fibrations, we find that a lift exists, which implies that the two maps ${B \rightarrow B}$ are indeed (right) homotopic. Thus ${i}$ is a homotopy equivalence, so an isomorphism in the homotopy category; it is in particular a weak equivalence by the next lemma.

2. A converse to the Whitehead theorem

We now need to show that a homotopy equivalence in a model category is a weak equivalence. While this is diagram-chasing, I don’t think it was displayed as prominently as the usual Whitehead theorem (that a weak equivalence between cofibrant-fibrant objects is a homotopy equivalence).

Lemma 7 If ${e: C \rightarrow C}$ is a map in a model category which is left-homotopic to the identity, then ${e}$ is a weak equivalence.

Proof: Indeed, let ${\mathrm{Cyl}(C)}$ be a cylinder object of ${C}$, so there is a morphism

$\displaystyle H:\mathrm{Cyl}(C) \rightarrow C$

which is a homotopy between the map ${e}$ and the identity. There are inclusions ${i_0, i_1: C \hookrightarrow \mathrm{Cyl}(C)}$ (which are cofibrations if ${C}$ is cofibrant) such that ${H i_0 = e, Hi_1 = 1_C}$. Now both ${i_0, i_1}$ are weak equivalences; two-out-of-three applied to the second identity shows that ${H}$ is a weak equivalence. Now two-out-of-three again implies that ${c}$ is a weak equivalence. We need the following brief lemma.

Lemma 8 In any model category, a map ${f: A \rightarrow B}$ that induces an isomorphism in the homotopy category is a weak equivalence.

Proof: Without loss of generality, we can assume that ${A, B}$ are cofibrant and fibrant, by replacing them with better objects. Then we know that the maps in the homotopy category between ${A, B}$ are just the homotopy classes of maps from ${A \rightarrow B}$.

So we have a homotopy equivalence

$\displaystyle f: A \rightarrow B$

between cofibrant-fibrant objects. We need to show that it is a weak equivalence. We can factor ${f}$ as

$\displaystyle A \stackrel{i}{\rightarrow} C \stackrel{p}{\rightarrow} B$

where ${i}$ is an acyclic cofibration and ${p}$ is a fibration. Clearly ${C}$ is cofibrant-fibrant as well. Now ${i}$ is a homotopy equivalence by the Whitehead theorem, so ${p}$ must be a homotopy equivalence too. We are reduced to showing that ${p}$ is a weak equivalence. In particular, we need only prove the result for fibrations.

Now let ${\mathrm{Cyl}(C)}$ be a cylinder object for ${C}$. There is a map ${q: B \rightarrow C}$ which is a homotopy inverse to ${p}$. We want a refinement of this map that will also be a section of ${p}$. So there is a homotopy ${H:\mathrm{Cyl}(B) \rightarrow B}$ between ${pq}$ and the identity. There are two inclusions ${i_0, i_1: B \hookrightarrow \mathrm{Cyl}(B)}$; let ${i_1}$ be the one corresponding to the identity in the above homotopy.

We get a diagram

There is a lift ${G: \mathrm{Cyl}(B) \rightarrow C}$ as ${p}$ is a fibration. Let ${\overline{q}= G i_1}$; then ${\overline{q}: B \rightarrow C}$ is homotopic to ${q}$, and is another homotopy inverse for ${p}$. But ${\overline{q}}$ has a redeeming feature that ${q}$ need not have; we have that ${p \circ \overline{q} = H \circ i_1 = 1_B}$.

So we are in the following situation. We have a fibration ${p: C \rightarrow B}$ between cofibrant-fibrant objects with a homotopy inverse ${\overline{q}: B \rightarrow C}$. We also have that ${p \circ \overline{q} = 1_B}$. We are to show that ${p}$ is a weak equivalence.

Now ${\overline{q} \circ p}$ is a weak equivalence by the previous lemma, as it is homotopic to the identity. Moreover, there is a diagram

Here the horizontal maps on the top row are the identities ${1_C}$. This commutes because ${p \circ \overline{q} = 1_B}$. This is a retract diagram, and we find that ${p}$ is a retract of the weak equivalence ${\overline{q} \circ p}$; it is thus a weak equivalence.

3. Simplicial algebras

We shall now use the fact that there is a cofibrantly generated model structure on the category of simplicial sets, as stated above. Let ${A}$ be any ring. Note that there is an adjunction between simplicial ${A}$-algebras (given by the forgetful functor and the free functor) and simplicial sets. It is a basic fact that every simplicial group (in particular, simplicial ring) is fibrant. As a simple corollary, we find:

Theorem 9 There is a cofibrantly generated model structure on the category ${\mathbf{S}\mbox{--}\mathbf{Alg}^A}$ of simplicial ${A}$-algebras. Generating cofibrations are given by ${Free(S_\bullet) \rightarrow Free(S'_\bullet)}$ where ${S_\bullet \rightarrow S'_\bullet}$ is an inclusion of simplicial sets; the small subset where ${S = \partial \Delta[n]_\bullet, S' = \Delta[n]_\bullet}$ suffices. A map is a fibration or weak equivalence in ${\mathbf{S}\mbox{--}\mathbf{Alg}^A}$ iff it is so in ${\mathbf{SSet}}$.