Homotopy on Chain Complexes

Let $ R$ be any associative ring with unit and $ A$ an $ R$-module. If $ P$ is a projective module and $ A\to P\to A = 1_A$, is $ A$ necessarily projective?

The answer is yes, as the reader probably surmised. The argument is short: suppose $ B\to C$ is a surjective homomorphism and $ A\to C$ is any homomorphism. Then $ P\to A\to B$ lifts to $ P\to B$ and $ A\to P\to B$ is the appropriate lift of $ A\to C$.

A retraction of a homomorphism $ f:X\to Y$ is a morphism $ g:A\to B$ that fits into the commutative diagram

$ \begin{matrix} A & \to & X & \to & A\\ \downarrow & ~ & \downarrow & ~ & \downarrow\\ B & \to & Y & \to & B\end{matrix}$

where $ A\to X\to A = 1_A$ and $ B\to Y\to B = 1_B$.

Now, a modification of the $ A\to P\to A$ arugment with $ P$ projective shows that if $ X\to Y$ is injective with projective cokernel, then $ A\to B$ is also injective with projective cokernel.

Now, this argument applies not just to the category of $ R$-modules but also the category of nonnegative chain complexes of $ R$-modules to show that cofibrations (see next paragraph) are closed under retractions (actually, also all chain complexes, but later we will need nonnegative chain complexes).

So now we switch to nonnegative chain complexes. What is a cofibration anyway? We call a morphism $ f:X\to Y$ of chain complexes a cofibration if termwise, each $ f_n$ is injective with projective cokernel. We call $ f$ a fibration if $ f_n$ is surjective for $ n > 0$. Finally, we call $ f$ a weak equivalence if it induces an isomorphism on homology.

Of course, in the full subcategory of modules, where a module is embedded as a complex concentrated in degree zero, weak equivalences are actually isomorphisms.

It turns out that all three types of maps are closed under retractions (exercise!). This is the start to giving the category of nonnegative chain complexes something called a model category structure invented by Daniel Quillen, which will allow us to do homotopy with them.

Model Categories

What is a model category structure? Briefly, a model category is a category together with three classes of morphisms: cofibrations, fibrations, and weak equivalences that were modelled after cofibrations, fibrations, and weak equivalences (either inducing an isomorphism on homotopy groups or a homotopy equivalence, depending on the setting), which allow one to do a surprising amount of homotopy theory without actually ever thinking about topological spaces. These morphisms are required to satisfy various axioms that we will look at later.

The advantage of this viewpoint is that many categories (such as chain complexes of $ R$-modules and simplicial sets) can be given a model category structure by introducing the three classes of morphisms, and then all the results of this abstract homotopy theory that we know about model categories can be used to prove results and organise ideas in our given category.

Some immediate advantages of doing this are that some of the ideas in homological algebra become more transparent and easier to prove, and that we also get a more flexible theory of derived functors that does not rely on abelian categories.

The Axioms

We start with the axioms. Let $ \mathcal{C}$ be a category. A model category structure on $ \mathcal{C}$ is $ \mathcal{C}$ itself together the specification of three families of morphisms on $ \mathcal{C}$: fibrations, cofibrations, and weak equivalences, which satisfy the following axioms:

(M1) $ \mathcal{C}$ has limits and colimits (sometimes only finite (co)limits are used)
(M2) All three families are closed under retractions
(M3) If $ gh$ is a defined composition and any two out of three of $ \{ g,h,gh\}$ are a weak equivalence, then all three are weak equivalences
(M4) Any lifting problem

$ \begin{matrix}A & \to & X\\ \downarrow & (\nearrow) & \downarrow\\ B & \to & Y\end{matrix}$

(i.e. can the diagonal arrow be filled in?) where $ A\to B$ is a cofibration, $ X\to Y$ is a fibration, and at least one of these is a weak equivalance, can be solved.
(M5) Any morphism $ X\to Y$ can be factored $ X\to Z\to Y$ where $ X\to Z$ is a cofibration and $ Z\to Y$ is a fibration, where we can choose either the first or the second to be a weak equivalence.

These axioms are short, and allow us to do homotopy. In the next few posts, we will do at least the following:

  • Sketch that nonnegative chain complexes of $ R$-modules form a model category.
  • Argue for why this is a good way of doing homotopy.
  • Give some applications such as André-Quillen homology.

For instance, we will see how the model structure on $ R$-modules allows us to conclude that projective resolutions are unique up to (chain) homotopy equivalence without ever trying to use the comparison lemma in homological algebra directly, which should indicate that we are working with a decent set of axioms.

Leave a comment

Fields marked with * are required. LaTeX snippets may be entered by surrounding them with single dollar signs. Use double dollar signs for display equations.