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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.

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    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.2011-05-24

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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.

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    Actually, we can talk about length-$n$ zigzags and length-$n$ sequences of elementary zigzag equivalences in $(\mathcal{C},W)$ just fine. So we just need to say, for each $n \in \mathbb{N}$, for all $f \in \mathrm{mor}\mathcal{C}$, if there is a length-$n$ zigzag and two length-$n$ sequences of elementary zigzag equivalences exhibiting it as inverse to $f$ in $\mathcal{C}[W^{-1}]$, then $f \in W$. The other way would be more compelling if we could _explicitly_ describe the saturation of a finite subset of arrows in a finitely-presentable category.2015-09-20