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I'm a little confused about a proof in this note by H. Krause; see page 12, Prop. 3.5.1. Can you explain me how the $\beta_i$'s are determined? I can't catch how to choose them in order to coequalize both $\alpha_i'$ and $\alpha_i''$, and I also think there's a mistake in the criterion for a morphism to be in the saturation of $\Sigma$... I'll also welcome any other reference for this proposition.

Thanks in advance.

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I think I'm able to prove it in a much simpler way than Krause's one; I'm just looking for a check on your experience, and considering this result turns out to be quite useful... Thanks! –  tetrapharmakon Dec 10 '11 at 13:34
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1 Answer

up vote 12 down vote accepted

For context, the question is about the following statement:

Suppose $\mathcal{C}$ is a category with small coproducts and that $\Sigma$ is a left multiplicative subset closed under taking coproducts, that is to say $\Sigma$ satisfies:

  1. (Nontriviality and closure under composition): $\Sigma$ contains all identity morphisms and is closed under taking compositions.
  2. (Ore condition): If $X'\; \xleftarrow{\sigma} \; X \; \xrightarrow{\alpha} \; Z$ is such that $\sigma \in \Sigma$ then there exists a commutative square
    Ore condition
    with $\sigma' \in \Sigma$.
    Note: If in addition $\alpha \in \Sigma$ then the diagonal $\sigma'\alpha: X \to Y'$ is a morphism in $\Sigma$ by 1.
  3. (Cancellation condition) If $\alpha,\beta: X \to Y$ are parallel morphisms for which there exists $\sigma: X' \to X$ in $\Sigma$ such that $\alpha\sigma = \beta\sigma$ then there exists $\sigma': Y \to Y'$ in $\Sigma$ such that $\sigma'\alpha = \sigma'\beta$
  4. (Closure under coproducts): If $\sigma_i : X_i \to Y_i$ belong to $\Sigma$ then so does $\coprod \sigma_i : \coprod X_i \to \coprod Y_i$.

Then the category of left fractions $\mathcal{C}[\Sigma^{-1}]$ has coproducts and the quotient functor $q: \mathcal{C} \to \mathcal{C}[\Sigma^{-1}]$ preserves coproducts.

The thing that needs to be shown is that the canonical map $$\tag{3.5.1} \mathcal{C}[\Sigma^{-1}]\left( \coprod X_i, Y\right)\; \xrightarrow{f}\; \prod\mathcal{C}[\Sigma^{-1}]\left( X_i, Y \right) $$ is a bijection. Krause first establishes surjectivity of $f$ and the point of contention is its injectivity.

I'll first fill in the details in Krause's argument (which is perfectly okay) because I think that nothing but straightforward manipulations of fractions is needed and getting used to those is a necessity if one wants to understand any of this.

1. Constructing the $\beta_i$'s

Given two left fractions $\coprod X_i \;\xrightarrow{\alpha^\prime} \; Z^\prime \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^\prime}\; Y$ and $\coprod X_i \;\xrightarrow{\alpha^{\prime\prime}} \; Z^{\prime\prime} \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^{\prime\prime}}\; Y$ that are identified under $f$ we want to show that the fractions are equivalent.

To say that the images of these two fractions under $f$ are identified is to say that for each $i$ the fractions $X_i \;\xrightarrow{\alpha_{i}^\prime} \; Z^\prime \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^\prime}\; Y$ and $X_i \;\xrightarrow{\alpha_{i}^{\prime\prime}} \; Z^{\prime\prime} \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^{\prime\prime}}\; Y$ are equivalent.

Notice that the diagram $Z^{\prime\prime} \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^{\prime\prime}}\; Y \; \xrightarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^{\prime}}\;Z'$ appears in all the fractions for all $i$. Applying the Ore condition to this wedge we get a morphism $Y \; \xrightarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma}\; Z$ of the form $\sigma= \tau' \sigma' = \tau''\sigma'' \in \Sigma$.

The diagram

Ore condition for the fractions

exhibits the fraction $ X_{i} \; \xrightarrow{\tau'\alpha_{i}'} \; Z \; \xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma} \; Y $ as equivalent to $ X_i \;\xrightarrow{\alpha_{i}^\prime} \; Z^\prime \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^\prime}\; Y $. Similarly, $ X_{i} \; \xrightarrow{\tau''\alpha_{i}^{\prime\prime}} \; Z \; \xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma} \; Y $ is equivalent to $ X_i \;\xrightarrow{\alpha_{i}^{\prime\prime}} \; Z^{\prime\prime} \;\xleftarrow[\vphantom{\ }^{\Large{\sim}}]{\sigma^{\prime\prime}}\; Y $.

At this point Krause replaces $\tau'\alpha_{i}^\prime$ by $\alpha_{i}^\prime$ and similarly for $\alpha_{i}^{\prime\prime}$ — this is justified by his sentence “We may assume that $Z^\prime = Z = Z^{\prime\prime}$ and $\sigma^\prime = \sigma = \sigma^{\prime\prime}$ since […]” — to avoid confusion we won't do this here.

Since equivalence of fractions is transitive (it is an equivalence relation — check this!), we conclude that the two fractions involving $Z$ are equivalent. But this means that we have a commutative diagram

Equivalence of fractions

From $\beta_{i}^\prime\sigma = \beta_{i}^{\prime\prime}\sigma$ we conclude with the cancellation condition that there is $\sigma_{i}^\prime: \tilde{Z}_i \to Z_i$ in $\Sigma$ such that $\sigma_{i}^\prime \beta_{i}^\prime = \sigma_{i}^{\prime}\beta_{i}^{\prime\prime}$. Now put $\beta_i = \sigma_{i}^\prime \beta_{i}^\prime = \sigma_{i}^{\prime}\beta_{i}^{\prime\prime}$ and observe that $\beta_{i} \tau'\alpha_{i}^\prime = \beta_{i}\tau^{\prime\prime}\alpha_{i}^{\prime\prime}$, as well as $\beta_i \sigma = \sigma_{i}^\prime (\beta_{i}^\prime \sigma) \in \Sigma$, as claimed by Krause (remembering the paragraph in italics above).

Added: To see that $\beta_i$ is indeed in the saturation of $\Sigma$ (see also the next section of the answer), we need only show that we can postcompose it with a morphism $\phi_i$ such that $\phi_i\beta_i \in \Sigma$. Here's one (not particularly elegant) way to do it:

Apply the Ore condition to the wedge $Z \; \xleftarrow{\sigma} \; Y \; \xrightarrow{\sigma_{i}^\prime \tilde{\sigma}_i} \; Z_i$ to get the commutative rectangle on the left below

Beta is in the saturation

The dotted diagonal arrow can be either one of $\beta_{i}^\prime$ or $\beta_{i}^{\prime\prime}$ and it doesn't make the diagram commutative. Nevertheless, $\tau_i \beta_i = \tau_i \sigma_{i}^\prime\beta_{i}^\prime$ and $\upsilon_i$ are two morphisms $Z \to \tilde{W}_i$ such that $\tau_i \beta_i\sigma = \upsilon_i \sigma$. Thus, by cancellation, we can find $\tau_{i}^\prime: \tilde{W}_i \to W_i$ in $\Sigma$ such that $\tau_{i}^\prime\tau_i\beta_i = \tau_{i}^\prime \upsilon \in \Sigma$ and thus $\phi_i = \tau_{i}^\prime \tau_i$ does what we want.

2. The criterion for a morphism to be in the saturation of $\Sigma$.

By definition, a morphism $\phi \in \mathcal{C}$ belongs to the saturation $\overline{\Sigma}$ of $\Sigma$ if and only if the localization functor $q: \mathcal{C} \to \mathcal{C}[\Sigma^{-1}]$ sends $\phi$ to an invertible morphism.

Claim. A morphism $\phi$ is in the saturation of $\Sigma$ if and only if there are morphisms $\phi'$ and $\phi''$ such that $\phi'' \phi \in \Sigma$ and $\phi \phi' \in \Sigma$.

Indeed, assume there are such $\phi'$ and $\phi''$. Then $(\phi''\phi)^{-1} \phi''$ is a left inverse of $\phi$ and $\phi'(\phi\phi')^{-1}$ is a right inverse of $\phi$ in $\mathcal{C}[\Sigma^{-1}]$, so $\phi$ is invertible, too.

Recall the explicit form of $q(\phi) = [\phi, 1]$. To say that $[\phi,1]$ has a left inverse is to say that there is a morphism $[\alpha,\sigma]$ such that $[\alpha,\sigma][\phi,1] \sim [1_A,1_A]$. Recalling the definition of the composition in the category of fractions we see that $[\alpha,\sigma][\phi,1] = [\alpha\phi,\sigma]$ and to say that the latter is equivalent to $[1,1]$ is to give a commutative diagram

Saturation argument

telling us that $(\beta \alpha)\phi = \sigma^\prime \in \Sigma$ so that we can take $\phi^{\prime\prime} = \beta\alpha$. Similarly for the proof of existence of $\phi^\prime$.

3. End of the argument

With these arguments at hand, I think it is safe to refer back to Krause's notes, which I reproduce for the convenience of the reader — (3.5.1) is the same as the displayed equation above:

End of the proof

4. Alternative proof and some references

The classic reference for the calculus of fractions is Gabriel-Zisman, Calculus of Fractions and Homotopy Theory, but it is relatively terse. A detailed exposition can be found in Schubert, Categories, chapter 19 (at least in the German edition). For those who want to verify the basics I always recommend having pages 300 ff. of Lam's Lectures on modules and rings nearby because the arguments given there can be easily generalized from rings to categories by keeping track of source and target of the morphisms.

As for an alternative proof of the proposition you ask about, you can appeal to the fact that the morphism classes in $\mathcal{C}[\Sigma^{-1}]$ can be described as filtered colimits and keeping track of those. This is done in the proof of Theorem 19.2.8 of (the German edition of) Schubert's book. Filling in the details doesn't seem easier to me than Krause's argument, though, as the verification of all the naturality assertions involves precisely the same arguments, but hidden in a formalism. However, this is probably a matter of taste…

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Thanks a lot! I spent some time in reading your answer in order to give you a sensible comment. Maybe you reversed the order of the composition $\tau'\alpha',\tau''\alpha''$ when you composed with $\beta_i$; in the second equivalence of squares, $\beta_i'\tau_i'\alpha_i'=\tau_i'\alpha_i'$ is clearly a mistake. Finally I can't see where is a $\tau$ such that $\tau\beta_i\in \Sigma$. Thanks for your patience and complete answer! –  tetrapharmakon Dec 19 '11 at 18:23
You're absolutely right about the reversal of the order and the other typo. I'll fix it and address your other question tomorrow. It's getting too late now. I'll ping you when it's done. –  t.b. Dec 19 '11 at 21:21
@tetrapharmakon: I've corrected a few typos and added an argument to the answer. I hope it's okay now. –  t.b. Dec 20 '11 at 1:40
Yeah, now it seems all right! Thanks! –  tetrapharmakon Dec 20 '11 at 5:57
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