# Isomorphisms in the localization of a category

Let $\mathcal{C}$ be a (small) category, and $S \subset \mathcal{C}$ a class of morphisms in $\mathcal{C}$. Suppose $f$ is a morphism in $\mathcal{C}$ that becomes an isomorphism in the localization $S^{-1}\mathcal{C}$. Suppose moreover that $S$ satisfies the two-out-of-three property (i.e. in a composition, if two of the terms belong to $S$, then so does the third) and contains all isomorphisms in $\mathcal{C}$. When can we conclude that $f \in S$ itself?

In the special case that I'm considering, $S$ is the class of weak equivalences in a model category $\mathcal{C}$. In this case it is true (and follows from the alternative description of the homotopy category) that an isomorphism in the homotopy category is a weak equivalence, but the proof involves some manipulations. I am curious if a simpler approach exists.

• I guess you don't want to have a description of the localization via a calculus of fractions (which as far as I know doesn't work for model categories) but in which stability of $S$ under weak pull-backs or push-outs is sufficient. There is a relatively recent preprint by Sebastian Thomas on various conditions on $S$ that you can impose, but I wouldn't call that simple and some manipulations would be a euphemism. The same can be said about the Dwyer-Hirschhorn-Kan-Smith approach, of which you certainly are aware and on which Thomas's paper is based. – t.b. May 24 '11 at 1:22
• @Theo: Dear Theo, thanks! I wasn't aware of any of this, though it does indeed appear somewhat difficult. – Akhil Mathew May 24 '11 at 2:20
• The calculus of fractions is due to Gabriel-Zisman and can be found (at least outlined) in Ch. 10 of Weibel. It is patterned after Ore's localization theory for non-commutative rings and gives you a handy description of the localized category. But the hypotheses are a bit too strong to apply to model categories. Nevertheless I'd have a look at that. It is treated nicely in an old book on categories by Schubert (Ch 19 of the German ed.). DHKS propose a localization approach for model categories based purely on weak equivalences and Thomas made a systematic study of all these theories. – t.b. May 24 '11 at 2:35
• @Theo: Thanks for the references! I hadn't considered that there was an analogy between Ore and G-Z localization; interesting. I'll take a look at ch. 10 of Weibel. – Akhil Mathew May 24 '11 at 14:32
• One last remark that may or may not be helpful: While checking that everything works fine in the category defined by fractions, I found pages 300-302 of Lam's Lectures on modules and rings very useful for orientation. It is very easy to get lost in all those diagrams (e.g. while checking transitivity of the equivalence relation or associativity of the composition) and Lam's exposition can be easily read to apply to categories, not only rings. – t.b. May 24 '11 at 15:32

## 1 Answer

Let $\mathcal{C}$ be a category, let $S \subseteq \operatorname{mor} \mathcal{C}$, and let $\bar{S}$ be the class of all morphisms in $\mathcal{C}$ that become invertible in $S^{-1} \mathcal{C}$. As t.b. pointed out in the comments, the 2-out-of-6 property + a three-arrow calculus is enough to guarantee $S = \bar{S}$, i.e. that the morphisms in $\mathcal{C}$ that become invertible in $S^{-1} \mathcal{C}$ are precisely the ones in $S$. (Note that the 2-out-of-6 property is a necessary condition).

If you are willing to take the fundamental theorem of three-arrow calculi (i.e. the one that gives necessary and sufficient conditions for two three-arrow zigzags (i.e. $\bullet \leftarrow \bullet \rightarrow \bullet \leftarrow \bullet$) to represent the same morphism) for granted, this is actually quite straightforward: see e.g. Proposition 36.4 in [Dwyer, Hirschhorn, Kan, and Smith] or proposition 3.5.10 in my notes.

The difficulty in the general case seems to boil down to the fact that the zigzag representing an inverse in $S^{-1} \mathcal{C}$ for a morphism in $\bar{S}$ may not consist of only morphisms in $\bar{S}$ (let alone $S$). A trivial example of this is the case where $S = \emptyset$ or $S = \{ \text{identities} \}$. However, observe that a three-arrow zigzag that represents an isomorphism in $S^{-1} \mathcal{C}$ necessarily consists of only morphisms in $\bar{S}$. This, I suppose, is the significance of 3.

Rather curiously, the fact that $\bar{S}$ is closed under retracts seems to play no role. For the record, let me point out that the 2-out-of-6 property does not imply closure under retracts. Consider the following category $\mathcal{C}$, $$\begin{array}{ccccc} X' & \to & X & \to & X' \\ \downarrow & & \downarrow & & \downarrow \\ Y' & \to & Y & \to & Y' \end{array}$$ where the composite across the top row is $\mathrm{id}_{X'}$ and the composite across the bottom row is $\mathrm{id}_{Y'}$, but $X' \to X$ and $Y' \to Y$ are not isomorphisms. Let $S$ be the set of all identity morphisms in $\mathcal{C}$, plus the morphism $X \to Y$. Then $S$ has the 2-out-of-6 property (because none of the morphisms in $S$ admit any non-trivial factorisation) but is not closed under retracts.

There is a small sliver of hope, though. Observe that the class of pairs $(\mathcal{C}, S)$ where $S = \bar{S}$ is closed under arbitrary products. (See lemma 3.1.11 in my notes.) Let us say that $(\mathcal{C}, S)$ is saturated if $S = \bar{S}$. The functor $(\mathcal{C}, S) \mapsto S^{-1} \mathcal{C}$ is a left adjoint, so it preserves colimits. In particular, it preserves filtered colimits. Moreover, given a small filtered diagram $\mathcal{A}_\bullet : \mathcal{J} \to \mathbf{Cat}$, a morphism in ${\varinjlim}_\mathcal{J} \mathcal{A}_\bullet$ is an isomorphism if and only if it is the image of an isomorphism in some $\mathcal{A}_j$; thus, filtered colimits preserve the property of being saturated. Hence, the class of pairs $(\mathcal{C}, S)$ where $S = \bar{S}$ is closed under ultraproducts. It is not hard to see that $(\mathcal{C}, S)$ is saturated if and only if some ultrapower is saturated, so the Keisler–Shelah theorem implies the class of saturated $(\mathcal{C}, S)$ is closed under elementary equivalence. It is therefore an elementary class, i.e. axiomatisable by a theory in the first-order language of categories with an extra unary predicate. Perhaps someone clever will be able to find an explicit description of this theory.

• I don't think you need the Keisler-Shelah theorem here: an inverse to $f$ in $\mathcal{C}[W^{-1}]$ is always witnessed by a (finite) zigzag and two (finite) equivalences of zigzags, so $W$ is saturated in $\mathcal{C}$ if and only if for every finite subcategory $\mathcal{D} \subseteq \mathcal{C}$, $\mathcal{D} \cap W$ is saturated in $\mathcal{D}$. So to write down the theory of saturation, you "just" need to compute the saturation of every (finite category, subcategory) pair. If the class of categories involved could be cut down further, that would be interesting. – tcamps Sep 20 '15 at 16:06
• It's not clear to me whether saturation is inherited by subcategories. – Zhen Lin Sep 20 '15 at 16:17
• $f$ lies in the saturation of $\mathcal{D} \cap W$ if and only if there is a zigzag $Z$ in $\mathcal{D}$ with the backward arrows in $\mathcal{D} \cap W$ and two equivalences of zig zags in $\mathcal{D}$ ensuring that $Z$ is inverse to $f$ in $\mathcal{D}[(\mathcal{D} \cap W)^{-1}]$. Then $Z$ is a zigzag in $\mathcal{C}$, and the equivalences witness that it is inverse to $f$ in $\mathcal{C}[W^{-1}]$. Am I missing something? – tcamps Sep 20 '15 at 16:21
• Right, sorry. I was thinking of a different issue. (It is not obvious whether or not a morphism in $\mathcal{D}$ can be in the saturation of $\mathcal{C} \cap \mathcal{W}$ without being in the saturation of $\mathcal{D} \cap \mathcal{W}$.) – Zhen Lin Sep 20 '15 at 16:30
• Right, yeah, that's surely possible. The witness data to lying in the (saturation of $\mathcal{W}$) $\cap \mathcal{D}$ need not lie in $\mathcal{D}$. You need to pass to a larger (but still finite!) subcategory to find it. And if you want $\mathcal{D}'$ with (saturation of $\mathcal{W}$ in $\mathcal{C}$) $\cap \mathcal{D}'$ = (saturation of $\mathcal{W}$ in $\mathcal{D}'$), you'll need to iterate this process, and $\mathcal{D}'$ won't typically be finite, but luckily we don't need this. – tcamps Sep 20 '15 at 16:39