## category fibered in groupoids

06/12/2020 Uncategorized

) A cartesian section is thus a (strictly) compatible system of inverse images over objects of E. The category of cartesian sections of F is denoted by, In the important case where E has a terminal object e (thus in particular when E is a topos or the category E/S of arrows with target S in E) the functor. In this section we explain how to think about categories fibred in groupoids and we see how they are basically the same as functors with values in the $(2, 1)$-category of groupoids. A category over Cis a category Fwith a functor p: F!C. C Thus a cartesian section consists of a choice of one object xS in FS for each object S in E, and for each morphism f: T → S a choice of an inverse image mf: xT → xS. The functor $\Delta _ G$ maps the object $x$ of $\mathcal{S}_ U$ to the triple $(x, x, \text{id}_{G(x)})$. For an example, see below. {\displaystyle s:G\times X\to X} As an application we obtain a Tannakian interpretation for the Nori fundamental gerbe defined in [BV] for non smooth non pseudo-proper algebraic stacks. c Its $2$-morphisms $t : G \to H$ for $G, H : (\mathcal{S}, p) \to (\mathcal{S}', p')$ will be morphisms of functors such that $p'(t_ x) = \text{id}_{p(x)}$ for all $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$. Lemma 4.35.11. Let $p : \mathcal{S} \to \mathcal{C}$ be a functor. Let $p : \mathcal{S} \to \mathcal{C}$ be a functor. Proof. More precisely, if φ: F →E is a functor, then a morphism m: x → y in F is called co-cartesian if it is cartesian for the opposite functor φop: Fop → Eop. Definition 4.35.6. Let $\mathcal{C}$ be a category. The tag you filled in for the captcha is wrong. This provides us with a general and flexible framework to study quantum field theories defined on spacetimes with extra geometric structures such as bundles, connections and spin structures. Let $\mathcal{C}$ be a category. Hence there exists an isomorphism $\alpha : x' \to G(f^*x)$ such that $p'(\alpha ) = \text{id}_{U'}$ (this time by the axioms for $\mathcal{S}'$). Fibred category Fibred categories (or fibered categories) are abstract entities in mathematics used to provide a general framework for descent theory.They formalise the various situations in geometry and algebra in which inverse images (or pull-backs) of objects such as vector bundles can be defined. ... can have the same cohomology, if the groupoids they represent are equivalent or even locally equivalent (in the at topology). → Let $\mathcal{C}$ be a category. s If $p : \mathcal{S} \to \mathcal{C}$ is a category fibred in groupoids and $p$ factors through $p' : \mathcal{S} \to \mathcal{C}/U$ then $p' : \mathcal{S} \to \mathcal{C}/U$ is fibred in groupoids. The $2$-category of categories fibred in groupoids over $\mathcal{C}$ is the sub $2$-category of the $2$-category of fibred categories over $\mathcal{C}$ (see Definition 4.33.9) defined as follows: If F is a fibred E-category, it is always possible, for each morphism f: T → S in E and each object y in FS, to choose (by using the axiom of choice) precisely one inverse image m: x → y. is a pullback square. for $x, y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}')$ the set of morphisms between $x$ and $y$ in $\mathcal{S}'$ is the set of strongly cartesian morphisms between $x$ and $y$ in $\mathcal{S}$. Consider the commutative diagram. : F However, we will argue using the criterion of Lemma 4.35.2. Let $\mathcal{C}$ be a category. {\displaystyle {\mathcal {F}}} from A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). X is fully faithful (Lemma 5.7 of Giraud (1964)). Suppose that $\varphi : \mathcal{S}_1 \to \mathcal{S}_2$ and $\psi : \mathcal{S}_3 \to \mathcal{S}_4$ are equivalences over $\mathcal{C}$. Let $p : \mathcal{S}\to \mathcal{C}$ and $p' : \mathcal{S'}\to \mathcal{C}$ be categories fibred in groupoids. The following are equivalent. Lemma 4.35.16. Note the fiber category over an object is just the associated groupoid from the original groupoid in sets. y c If $\mathcal{A}$ is fibred in groupoids over $\mathcal{B}$ and $\mathcal{B}$ is fibred in groupoids over $\mathcal{C}$, then $\mathcal{A}$ is fibred in groupoids over $\mathcal{C}$. fully faithful) we have to show for any objects $x, y\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$ that $G$ induces an injection (resp. Poisson manifolds and their associated stacks Co-author(s): -Status: Published in Letters in Mathematical Physics Abstract: We associate to any integrable Poisson manifold a stack, i.e., a category fibered in groupoids over a site. Let $\mathcal{S}_ i$, $i = 1, 2, 3, 4$ be categories fibred in groupoids over $\mathcal{C}$. Two Examples of Integrable Category Fibered in Groupoids In the present x1, we give two examples of integrable [cf. (By the second axiom of a category fibred in groupoids.) PDF | We introduce an abstract concept of quantum field theory on categories fibered in groupoids over the category of spacetimes. . ( We introduce an abstract concept of quantum field theory on categories fibered in groupoids over the category of spacetimes. }function () { Lemma 4.35.14. a [2], Deﬁnition 1.7] categories ﬁbered in groupoids [cf. , and using the Grothendieck construction, this gives a category fibered in groupoids over This provides us with a general and flexible framework to study quantum field theories defined on spacetimes with extra geometric structures such as bundles, connections and spin structures. fully faithful) gives the desired result. By Lemma 4.32.4 it is enough to show that the $2$-fibre product of groupoids is a groupoid, which is clear (from the construction in Lemma 4.31.4 for example). a Lemma 4.35.8. There can in general be more than one cartesian morphism projecting to a given morphism f: T → S, possibly having different sources; thus there can be more than one inverse image of a given object y in FS by f. However, it is a direct consequence of the definition that two such inverse images are isomorphic in FT. A functor φ: F → E is also called an E-category, or said to make F into an E-category or a category over E. An E-functor from an E-category φ: F → E to an E-category ψ: G → E is a functor α: F → G such that ψ ∘ α = φ. E-categories form in a natural manner a 2-category, with 1-morphisms being E-functors, and 2-morphisms being natural transformations between E-functors whose components lie in some fibre. Then $p' : \mathcal{S}' \to \mathcal{C}$ is fibred in groupoids. ( Ob Let $\mathcal{C}$ be a category. c Abstract: We introduce an abstract concept of quantum field theory on categories fibered in groupoids over the category of spacetimes. / One example is the functor from Example 4.35.4 when $G \to H$ is not surjective. t is an equivalence. The main application of fibred categories is in descent theory, concerned with a vast generalisation of "glueing" techniques used in topology. $\square$. $p : \mathcal{S} \to \mathcal{C}$ is a category fibred in groupoids, and {\displaystyle a:G\to {\text{Aut}}(X)} The choice of a (normalised) cleavage for a fibred E-category F specifies, for each morphism f: T → S in E, a functor f*: FS → FT: on objects f* is simply the inverse image by the corresponding transport morphism, and on morphisms it is defined in a natural manner by the defining universal property of cartesian morphisms. C an equivalence) if and only if for each $U\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ the induced functor $G_ U : \mathcal{S}_ U\to \mathcal{S}'_ U$ is faithful (resp. Now fix $f : U \to V$. given by. Denote $p : \mathcal{X} \to \mathcal{C}$ and $q : \mathcal{Y} \to \mathcal{C}$ the structure functors. We will show that both $\mathcal{X}'$ and $\mathcal{X}''$ over $\mathcal{Y}$ are equivalent to the category fibred in groupoids $\mathcal{X} \times _{F, \mathcal{Y}, \text{id}} \mathcal{Y}$ over $\mathcal{Y}$, see proof of Lemma 4.35.15. $\mathop{\mathrm{Ob}}\nolimits (\mathcal{S}') = \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$, and Definition 0.3. p The set of morphisms from $x$ to $y$ lying over $f$ is bijective to the set of morphisms between $x$ and $f^*y$ lying over $\text{id}_ U$. To do this we argue as in the discussion following Definition 4.35.1. Lemma 4.35.15. : Suppose that $x'$ lies over $U'$ and $x$ lies over $U$. 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