Antworte auf:  Äquivalente Definitionen eines Stacks von xiao_shi_tou_
Forum:  Kategorientheorie, moderiert von: Buri Gockel

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xiao_shi_tou_
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Dabei seit: 12.08.2014
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 Beitrag No.10, eingetragen 2019-11-19 17:47    [Diesen Beitrag zitieren]

Perfekt!
Vielen Dank.
Die Notizen hier die ich momentan lese sind wohl nicht ganz fehlerfrei, aber ich weiß jetzt, wie gemacht werden soll :D.

Vielen Dank für deine Geduld mit mir.



Triceratops
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 Beitrag No.9, eingetragen 2019-11-19 17:42    [Diesen Beitrag zitieren]

Ja, du hast recht.

Man muss allerdings noch verlangen, dass kartesische Pfeile erhalten bleiben:

Definition 3.6. in Notes on Grothendieck topologies,fibered categories and descent theory von Vistoli.


xiao_shi_tou_
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 Beitrag No.8, eingetragen 2019-11-19 17:12    [Diesen Beitrag zitieren]
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2019-11-19 15:06 - Triceratops in Beitrag No. 7 schreibt:
Nun man möchte schon für alle $U \in \mathcal{C}$ nicht nur eine Abbildung $\alpha(U) : \mathcal{F}(U) \to \mathcal{G}(U)$, sondern einen Funktor haben. Und für Morphismen $f : U \to V$ muss man einen Isomorphismus $f^* \circ \alpha(V) \to \alpha(U) \circ f^*$ von Funktoren $\mathcal{F}(V) \to \mathcal{G}(U)$ haben.
Hallo.
Hm, schränkt sich der Funktor $\a\colon \c{F}\to \c{G}$ nicht einfach zu einem Funktor auf den Fasern ein indem man die Inklusion $\c{F}\mid_U\sube \c{F}$ vorschaltet?
Ich glaube du beziehst dich auf die Definition eines Stacks als Funktor
$\c{F}\colon \c{C}^{\o{op}}\to \mathbf{Grpd}$, demnach wären Morphismen Morphismen von $2$-Funktoren.

Ich meinte aber einen Morphismus von Stacks als gefaserte Kategorien.
Bei gefaserten Kategorien sehe ich nicht ein, warum man hier einen $2$-Funktor zwischen den gefaserten Kategorien haben sollte, zumal die gefaserten Kategorien selbst gewöhnliche Kategorien sind.

Viele Grüße
EDIT:
Es ist jetzt klar:

Fasst man stacks als gefaserte Kategorien auf $\arr{\c{F}}{\rho_{\c{F}}}{\c{C}}$, dann sind die Morphismen Morphismen von gefaserten Kategorien, also gewöhnliche Funktoren
$$\a\colon \c{F}\to \c{G}$$, sodass $\rho_{\c{F}}=\rho_{\c{G}}\circ \a$, sodass stark kartesische Morphismen (strongly cartesian morphisms) auf stark kartesische Morphismen geschickt werden.


Fasst man stacks als $2$-Funktoren auf, dann sind die Morphismen Morphismen von $2$-Funktoren.
\(\endgroup\)

Triceratops
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 Beitrag No.7, eingetragen 2019-11-19 15:06    [Diesen Beitrag zitieren]

Nun man möchte schon für alle $U \in \mathcal{C}$ nicht nur eine Abbildung $\alpha(U) : \mathcal{F}(U) \to \mathcal{G}(U)$, sondern einen Funktor haben. Und für Morphismen $f : U \to V$ muss man einen Isomorphismus $f^* \circ \alpha(V) \to \alpha(U) \circ f^*$ von Funktoren $\mathcal{F}(V) \to \mathcal{G}(U)$ haben.


xiao_shi_tou_
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Dabei seit: 12.08.2014
Mitteilungen: 1250
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 Beitrag No.6, eingetragen 2019-11-19 14:17    [Diesen Beitrag zitieren]
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Hallo Triceratops.
Ich hätte da noch eine Frage zum Thema.

In den Notizen die ich lese wird ein Morphismus von stacks definiert als ein $2$-Funktor $f\colon \c{F}\to \c{G}$, welcher mit den Strukturmorphismen $\rho_{\c{F}}\colon \c{F}\to \c{C},\rho_{\c{G}}\colon \c{G}\to \c{C}$ kommutiert.
Hier sehe ich nicht ein, warum das ein $2$-Funktor sein soll, denn die gefaserten Kategorien $\c{F}$ und $\c{G}$ sind ja sogar gewöhnliche Kategorien. Ich nehme an das ist ein Tippfehler des Autors..!?

Viele Grüße
XST
\(\endgroup\)

xiao_shi_tou_
Senior
Dabei seit: 12.08.2014
Mitteilungen: 1250
Herkunft: Bonn
 Beitrag No.5, eingetragen 2019-11-19 02:34    [Diesen Beitrag zitieren]
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Vielen Dank für die hilfreichen Hinweise.
Die Konstruktion in hier ist demnach nicht ganz korrekt, aber nun ist alles klar wie man von einer zur anderen Definition übergeht und auch das mit dem $2$-Morphismus ist mir jetzt klar.



\(\endgroup\)

Triceratops
Aktiv
Dabei seit: 28.04.2016
Mitteilungen: 4570
Herkunft: Berlin
 Beitrag No.4, eingetragen 2019-11-19 00:20    [Diesen Beitrag zitieren]

Zu deinen Edits in Beitrag 0 (besser, weil auffälliger, wäre es gewesen, neue Posts zu machen):

Was du ganz am Ende schreibst, ist richtig.

Für einen $2$-Funktor $F : \mathcal{C}^{\mathrm{op}} \to \mathbf{Grpd}$ definiert man die Kategorie $\int F $ (Grothendieck-Konstruktion) so: Die Objekte sind Paare $(U,L)$, wobei $U \in \mathcal{C}$ und $L \in F(U)$. Ein Morphismus $(f,\alpha) : (U,L) \to (V,K)$ ist ein Morphismus $f : U \to V$ in $\mathcal{C}$ zusammen mit einem Isomorphismus $\alpha : f^*(K) \to L$ in $F(U)$, wobei $f^* := F(f)$. Die Komposition usw. sind leicht zu definieren. Der Funktor $\rho : \int F \to \mathcal{C}$ ist die Projektion auf die erste Komponente.

Behalte am besten immer Standardbeispiele (zum Beispiel der Stack der Geradenbündel; daher auch meine Notation oben) im Hinterkopf.

Was übrigens gerne verschwiegen wird: $\mathbf{Grpd}$ hat als Objekte lediglich die (essentiell) kleinen Gruppoide. Ansonsten bekommt man hier nämlich Probleme.


Triceratops
Aktiv
Dabei seit: 28.04.2016
Mitteilungen: 4570
Herkunft: Berlin
 Beitrag No.3, eingetragen 2019-11-19 00:16    [Diesen Beitrag zitieren]

Ja, genau.


xiao_shi_tou_
Senior
Dabei seit: 12.08.2014
Mitteilungen: 1250
Herkunft: Bonn
 Beitrag No.2, eingetragen 2019-11-18 23:57    [Diesen Beitrag zitieren]
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2019-11-18 23:26 - Triceratops in Beitrag No. 1 schreibt:
Die $2$-Kategorie $\mathbf{Grpd}$ hat nicht-triviale $2$-Morphismen, und die Isomorphismen $(f \circ g)^* \cong g^* \circ f^*$ sind in der Regel keine Identitäten.
 
Reicht das als Hinweis?

Verstehe.
In einer Gefaserten Kategorie über Gruppoiden gilt für den Pullback Funktor erstmal nur $(f\circ g)^*\cong g^*\circ f^*$ (also nur bis auf natürliche Isomorphismen $=$ $2$-Isomorphismen in $\mathbf{Grpd}$). Man bekommt also keinen Funktor, denn für diesen müsste $F(f\circ g)=F(g)\circ F(f)$ gelten und man definiert ja $F$ auf $1$-Morphismen als den Pullback Funktor, wenn man von Definition $1$ zu Definition $2$ übergeht.
\(\endgroup\)

Triceratops
Aktiv
Dabei seit: 28.04.2016
Mitteilungen: 4570
Herkunft: Berlin
 Beitrag No.1, eingetragen 2019-11-18 23:26    [Diesen Beitrag zitieren]

Die $2$-Kategorie $\mathbf{Grpd}$ hat nicht-triviale $2$-Morphismen, und die Isomorphismen $(f \circ g)^* \cong g^* \circ f^*$ sind in der Regel keine Identitäten.
 
Reicht das als Hinweis?


xiao_shi_tou_
Senior
Dabei seit: 12.08.2014
Mitteilungen: 1250
Herkunft: Bonn
 Themenstart: 2019-11-18 23:22    [Diesen Beitrag zitieren]
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Hallo zusammen.
Mir sind nun während meiner jungfräulichen Beschäftigung mit stacks zwei äquivalente Definitionen über den Weg gelaufen:
Man kann einen stack definieren, als Kategorie, welche in Gruppoiden gefasert ist mit Zusatzbedingungen:

Ein stack (in Gruppoiden) ist dann ein Funktor $\rho \colon \c{S}\to \c{C}$ wobei $\c{C}$ ein Situs ist (meistens ist das wohl der große $\etale$ situs $\mathbf{Sch}/S_{\et}$) zusammen mit Zusatzbedingungen (gefasert in Gruppoiden usw. )

Die andere Definitionsmöglichkeit ist einen stack zu definieren als einen $2$-Funktor $F\colon \c{C}^{\o{op}}\to \mathbf{Grpd}$, sodass sich Objekte in
$F(U)$ verkleben und Morphismen eindeutig verkleben lassen.

Aus welchem Grund verlangt man, dass $F$ ein $2$-Funktor ist (reicht hier ein Funktor gewöhnlicher Kategorien nicht aus?), bzw. wo wird das gebraucht um von der zweiten Definition zur ersten zu kommen?

Ich meine, die $2$-Kategorie $\c{C}$ ist doch sowieso strikt, also die $2$-Morphismen sind sowieso nur Identitäten. Kann man dann nicht jeden Funktor $\c{C}\to \mathbf{Grpd}$ kanonisch zu einem $2$-Morphismus "fortsetzen"?

Vielen Dank, ich bin gerade etwas verwirrt.

Vielleicht liegt meine Verwirrung daran, dass ich nicht verstehe, wie man von der $2.$ Definition zur ersten kommt:
In den Notizen die ich dazu lese wird folgendes geschrieben:
"Wenn $F\colon \c{C}\to \mathbf{Grpd}$ ein Stack im $2.$ Sinn ist, dann definiert man die Kategorie $\c{S}$ wie folgt:
Objekte sind Paare $(U,X)$ wo $U$ ein Schema ${}/S$ ist und $X$ ein Gruppoid. Ein Morphismus $(U,X)\to (U',X')$ soll aus einem Morphismus von Schemata $f\colon U\to U'$ sowie einem Isomorphismus $f^*X'\to X$ bestehen, wobei $f^*=\mathbf{Grpd}(f)$ ist."

Ich kann mir unter $\mathbf{Grpd}(f)$ nichts vorstellen, da das eine ein Morphismus von Schemata ist und das andere eine $2$-Kategorie.
Außerdem finde ich es komisch, dass man beliebige Gruppoide $X$ als Objekte zulässt. Ich hätte die Kategorie $\c{S}$ aus dem Bauch raus eher definiert als Paare $(U,F(U))$ mit Morphismen $\a\colon (U,F(U))\to(U',F(U'))$, wo $\a=(f,F(f))$ ist, also $f\colon U\to U'$ ein Morphismus von Schemata ${}/S$ und $F(f)\colon F(U')\to F(U)$ das Bild von $f$ unter $F$.
 
Viele Grüße
XST

EDIT:
Ich glaube es langsam zu verstehen:
Man nimmt als Objekte von $\c{S}$ Paare $(U,x)$ wo $x\in F(U)$. Dann ist ein Morphismus von $(U,x)$ nach $(U',x')$ ein Schemata Morphismus $f\colon U\to U'$, zusammen mit einem Isomorphismus $f^*x'\colon \defeq F(f)(x')\sto F(x)$.
Denn $\rho\colon \c{S}\to \c{C}$ muss ja als Fasern $F(U)$ haben und man definiert dann wohl $\rho$ durch $(U,x)\mapsto U$.


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