The Stacks project

94.4.0.1
\begin{equation} \label{examples-stacks-equation-morphisms-qcoh} \mathop{\mathrm{Mor}}\nolimits _{\mathcal{QC}\! \mathit{oh}}((Y, \mathcal{G}), (X, \mathcal{F})) = \coprod \nolimits _{f \in \mathop{\mathrm{Mor}}\nolimits _ S(Y, X)} \mathop{\mathrm{Mor}}\nolimits _{\mathit{QCoh}(\mathcal{O}_ Y)}(f^*\mathcal{F}, \mathcal{G}) \end{equation}

Comments (7)

Comment #3884 by Luke on

Typo: It should be instead of the other direction and the refering of f-maps is also unnecessary.

Comment #3928 by Luke on

Typo: It should be instead of the other direction and the refering of f-maps is also unnecessary.

Let me explain why I think the definition here is inconsistent. Certainly, this doesn't affect the propery of being stack. But following the current definition the fiber category over a scheme is rather than itself. If we see quasi coherent sheaves living in as in other chapters, then the pull back of along is . To be consistent with the definition of stack of algebraic spaces, I think it's more appropriate to turn arrows around.

Comment #3929 by on

Thanks very much. This is indeed a typo. I fixed it here. I think your explanation in the second paragraph of your second comment is just trying to convince me it was a typo? If there is something else wrong, then please let me know. Thanks again!

Comment #3946 by Luke on

Current fix is invalid. There are 3 places need to fix. First, I think it should be in this formula.

Second, the sentence below "See the discussion on f-maps of modules ..." should be deleted since is not the definition of f-maps.

Third, in proof of Lemma 04U3 below, it should be

and .

Comment #3947 by Luke on

The code can't display is Mor_{QCoh(\mathcal{O}_Z)} (\mathcal{H}, g^ f^\mathcal{F})

Comment #3953 by on

Dear Luke, I finally understand what you are saying. First of all, the thing you were pointing out wasn't a typo, it was actually correct, so I reverted the commit, see here. What you were really saying was (1) with the current definition the fibre categories of the stack QCoh are the opposite of the categories of quasi-coherent modules and (2) therefore we should reverse the arrows between the quasi-coherent sheaves everywhere. Now I agree with (1), good catch thanks!, and I fixed this here. But I don't agree with (2). I'll try to explain why.

The reason is that there is a duality between functions and spaces. If is a morphism of schemes, then we get on global sections. Now I think of a quasi-coherent -module as something you can turn into a vector bundle, see Section 27.6. And the stack of vector bundles Vect should have objects and morphisms consisting of pairs of maps and in the same direction. Thus, dually, for quasi-coherent modules the arrow should go in the opposite direction, which explains my choice in the text.

Now, when I first looked at Luke's suggestion, I wanted to say that you don't get a fibred category with Luke's choices. But when I worked it out, it turns out it works. I think there is a very general (somewhat boring) reason for this. I will explain this in the next paragraph, but please everybody: don't read this.

Suppose that we have a fibred category . Then for a morphism in we can choose a pullback functor on fibre categories, see Section 4.33. For simplicity, let's assume pullback functors compose on the nose. Then we can define a new category with the same objects as but morphisms are pairs where is a morphism in and in . There is a natural functor whose fibre categories are the opposites of the fibre categories of . Finally is a fibred category as well.

What I think is happening above. You can define my version of using the notion of -maps (you don't need to know about pullback and pushforward of sheaves in order to do this; this is silly but I am trying to make a point about what comes first so to speak). Then you can prove it is a fibred category whose pullback functors are the "usual" pullbacks. After you've done this you can construct the fibred category suggested by Luke by the procedure in the previous paragraph.

Comment #3962 by Luke on

Thanks for your reply!

Technically, there is no difference between these two definitions of , since weak 2-sheaf condition doesn't distinguish the functor and (which corresponding to your 4-th paragraph). So the choice of definition will mostly depend on philosophical issues( e.g. dulity of functions and spaces). My consideration is mainly from the functor of points POV. The followings are just for the sake of completeness.

This problem eventually reduce to the definition of vector bundles. Given a finite locally free sheaf over a scheme , both and are used by some authors to define the associated vector bundle over . Except that the first one is uesed in EGA II, I can't find any reason which shows that this is the canonical one.

First, it seems that Section 27.6 is mainly about (generalized) swan-serre theorem, i.e. Lemma 27.6.3. While the Serre-Swan theorem (especially Swan's) tells us that the equivalence is given by which maps a vector bundle to it's (sheaf of) sections, which is a covariant functor instead of the contravariant one in Lemma 27.6.3 there. This problem can be solved by using the second definition of associated vector bundle. To generalize serre-swan, one shouldn't expect to be an equivalence if is not.

Thus in here I can' see why the priciple of dulity of fuctions and spaces should come in. Rather, when talking about duality of functions and spaces, I think the ringed object such that can explain the f-maps as done in topology.

Personally, I think taking the second definition is natural and has technical advantage in the proof of Lemma 27.6.3. If we treat sheaves of modules as being induced from vector bundles by taking global sections as in Proposition 59.17.1, then the first difinition of associated vector bundle is inconsistent since and are generally different. From the functor of points POV, if we take the second definition, then the construction of associated vector bundle will become a concrete realization of the representable functor defined in Proposition 59.17.1 when given there is finite locally free.


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