Stacks on Geometry and some category theory

Seminar on Geometry/Topology of Principal/fiber bundles

Sun, 11 Aug 2019 14:13:32 +0000

Here I will add notes of the seminar that I am planning to conduct in School of Mathematics, IISER Thiruvananthapuram, India.

First lecture is expected to happen on 14 August 2019.
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Some terms which I want to convey the meaning of in this Seminar.

Manifold.
Differential forms on Manifolds; pullbacks and differential of a Differential form.
Lie group.
Lie algebra of Lie group.
Cohomology of Manifolds / Cohomology of Lie groups.
Principal/Vector bundle.
Connection (on principal/vector bundle).
Curvature (of Connection on principal/vector bundle).
Holonomy group.
Ambrose-Singer theorem.
Characteristic classes (Euler/Chern classes).

Lecture notes/Articles :

Books:

YouTube videos :

Fredric Schuller's YouTube channel

MathOverflow/MathStackExchange questions/user pages:

John M. Lee 's MathStackExchange page

Notation

Sat, 02 Feb 2019 17:26:24 +0000

Here I add notes about notation I use in this blog.

I might use the notion of fibered category and category fibered in groupoids as if there is no difference. I am mostly interested in fibered category $\mathcal{F}\rightarrow \mathcal{C}$ where the fibre $\mathcal{F}(U)$ is a groupoid for every object $U$ of $\mathcal{C}$. So, most of the times when I say fibered category, it is most likely that I mean fibred categroy whose fibres are groupoids i.e., category fibered in groupoids. Please let me know if there is some real confusion.

Stackification of fibred categories

Sat, 02 Feb 2019 13:50:32 +0000

I understood most of this from Introduction to the language of stacks and gerbes (section 2) by Ieke Moerdijk and from stacks project Stackification of fibred categories. It is necessary to know what is the sheafification of a presheaf to understand what is the stackification. I studied sheafification from Hartshorne's Algebraic geometry book. You can choose what you are comfortable with. I will mention the result first as in Lemma $8.8.1$.

Lemma : Let $\mathcal{C}$ be a site. Let $p:\mathcal{S}\rightarrow \mathcal{C}$ be a fibred category over $\mathcal{C}$. There exists a stack $p':\mathcal{S}'\rightarrow \mathcal{C}$ and a morphisms $G:\mathcal{S}\rightarrow \mathcal{S}'$ of fibred categories over $\mathcal{C}$ such that

for every $U\in \text{Ob}(\mathcal{C})$ and $x,y\in \mathcal{S}(U)$, the map $\text{Mor}(x,y)\rightarrow \text{Mor}(G(x),G(y))$ induced by $G$ identifies the right hand side with the sheafification of the left hand side.
For $U\in \mathcal{C}_0$ and $x'\in \mathcal{S}'(U)$ there exists a covering $\{U_i\rightarrow U\}$ such that each $x'|_{U_i}$ is in the essential image of the functor $G:\mathcal{S}(U)\rightarrow \mathcal{S}'(U)$.

We recall what is $\text{Mor}(a,b)$. This is a presheaf on $U$ defined as follows. Given an inclusion $i : V\hookrightarrow U$ we have $i^*(a),i^*(b)\in \mathcal{S}(V)$.

Category theory

Thu, 24 Jan 2019 23:24:08 +0000

In this page, I will give links of Category theory posts that I have made here. I learned some category theory from

Hilton and Stammbach's book A Course in Homological Algebra.
Angelo Vistoli's Descent theory notes.

Equivariant maps are Isomorphisms

Wed, 23 Jan 2019 17:49:33 +0000

Let $G$ be a Lie group and $\pi_P:P\rightarrow M, \pi_Q:Q\rightarrow M$ be principal $G$ bundles. Then, any $G$-equivariant map $f:P\rightarrow Q$ inducing identity on $M$ is a diffeomorphism. The same holds when we have Lie groupoids instead of Lie groups. Let $\mathcal{G}$ be a Lie groupoid and $P\rightarrow M, Q\rightarrow M$ be principal $\mathcal{G}$ bundles. Then, any $\mathcal{G}$-equivariant map $f:P\rightarrow Q$ inducing identity on $M$ is a diffeomorphism. Above result is very basic thing when defining a stack associated for a Lie groupoid $\mathcal{G}$. Given a Lie groupoid $\mathcal{G}$, we define a category fibered in groupoids $B\mathcal{G}\rightarrow \text{Man}$ by associating for each manifold $U$ a category $B\mathcal{G}(U)$ whose objects are principal $\mathcal{G}$ bundles whose base space is $U$ i.e., of the form $P\rightarrow U$ and morphism from an object $P\rightarrow U$ to another object $Q\rightarrow U$ is a $\mathcal{G}$-equivariant map $P\rightarrow Q$ that induces $Id:U\rightarrow U$ on base space of those principal bundles. Thus, to say $B\mathcal{G}(U)$ is a Lie groupoid, we need to prove that every arrow $(P\rightarrow U)\rightarrow (Q\rightarrow U)$ is an isomorphism which is what we are trying to prove. Let us see the proof for the case of Lie groups. See the set up as following diagram. Let $p,p'\in P$ are such that $f(p)=f(p')$, thus, $\pi_Q(f(p))=\pi_Q(f(p'))$. As $\pi_Q\circ f=\pi_P$, we have $\pi_P(p)=\pi_P(p')$ i.e., there exists $g\in G$ such that $p'=p.g$. Thus, $f(p')=f(pg)$. As $f$ is $G$-equivariant, we have $f(pg)=f(p)g$. Thus, we have $f(p')=f(p)g$. As the action of $G$ on $Q$ is free, $f(p')=f(p),f(p')=f(p)g$ implies $g=1$. Thus, $p'=p$. So, $f$ is one to one mapping. Let $q\in Q$. We have $\pi_Q(q)\in M$. As $\pi_P$ is surjective, there exists $p\in P$ such that $\pi_P(p)=\pi_Q(q)$. As $\pi_Q\circ f=\pi_P$, we have $\pi_Q(f(p))=\pi_P(p)=\pi_Q(q)$. As $\pi_Q(f(p))=\pi_Q(q)$, there exists $g\in G$ such that $f(p)g=q$. As $f$ is $G$-equivariant, we have $f(p)g=f(pg)$. Thus, we have $q=f(pg)$ which implies that $f$ is an onto mapping. Suppose that $\pi_P:P\rightarrow M$ is trivial $G$ bundle, not for simplicity but because every principal $G$ bundle is locally trivial and diffeomorphism is something that needs to be checked locally. As $\pi_P:P\rightarrow M$ is trivial, it has a global section for $\pi_P$ i.e., a smooth map $\sigma:M\rightarrow P$ such that $\pi_P\circ \sigma=1$. This gives a trivialization $M\times G\xrightarrow{\Phi} P$ i.e., an isomorphism. Consider the cimposition $f\circ \pi:M\rightarrow Q$. This is again a smooth map such that

$\pi_Q\circ (f\circ \sigma)=(\pi_Q\circ f)\circ \sigma=\pi_P\circ \sigma=1$

Definition of gerbe over stack

Sun, 20 Jan 2019 19:12:18 +0000

A morphism of stacks $F:\mathcal{D}\rightarrow \mathcal{C}$ is said to be a gerbe over stack if following two conditions hold :

Given a manifold $U$ and an object $\xi\in \mathcal{C}(U)$, there exists a covering $\{U_i\rightarrow U\}$ (depending on the Grothendieck topology that we have fixed on the category $Man$ of manifolds) and objects $x_i\in \mathcal{D}(U_i)$ with an isomorphism $F(x_i)\rightarrow \xi|_{U_i}$ for each $i$.
Given a manifold $U$ and an arrow $\xi\rightarrow \eta$ in $\mathcal{C}(U)$, there exists a covering $\{U_i\rightarrow U\}$ (depending on the Grothendieck topology that we have fixed on the category $Man$ of manifolds) and arrows $x_i\rightarrow y_i$ in $\mathcal{D}(U_i)$ such that

Lie groupoids

Wed, 16 Jan 2019 18:37:08 +0000

This post is based on (wanted to write after reading) Lie Groupoids and Differentiable stacks by Matias L. del Hoyo. I would suggest this for any one who wants to know about Lie groupoids and Differentiable stacks. This is well written. By a manifold, we always mean a smooth manifold. A Groupoid is a category where every arrow is invertible. A Lie groupoid is a groupoid with additional smooth structures on object set/morphism set and maps between them.

Definition : A Lie groupoid consists of a manifold $\mathcal{G}_0$ of objects, a manifold $\mathcal{G}_1$ of arrows and following maps :

$s:\mathcal{G}_1\rightarrow \mathcal{G}_0$, a submersion, called the source map.

$t:\mathcal{G}_1\rightarrow \mathcal{G}_0$, a submersion, called the target map.

$m:\mathcal{G}_1\times_{s,\mathcal{G}_0,t}\mathcal{G}_1\rightarrow \mathcal{G}_1$, a smooth map, called the multiplication map.

$u:\mathcal{G}_0\rightarrow \mathcal{G}_1$, a smooth map, called the unit map.

$i:\mathcal{G}_1\rightarrow \mathcal{G}_1$, a smooth map, called the inverse map.

with some compatibility conditions. We denote this Lie groupoid by $\mathcal{G}_1\rightrightarrows \mathcal{G}_0$. Definition : Let $\mathcal{G}_1\rightrightarrows \mathcal{G}_0$ be a Lie groupoid and $x\in \mathcal{G}_0$.

The set $s^{-1}(x)=:G(x,-)$ is called the $s$-fibre of $x$ .

The set $t^{-1}(x)=:G(x,-)$ is called the $s$-fibre of $x$

The set $s^{-1}(x)\cap t^{-1}(x)=:G_x$ is called the Isotropy group of $x$.

The set $t(s^{-1}(x))=\{y:x\rightarrow y\in \mathcal{G}_1\}=:O_x$ is called the orbit of $x$.

Proposition : Given a Lie groupoid $\mathcal{G}_1\rightrightarrows \mathcal{G}_0$ and $x,y\in \mathcal{G}_0$,

the subset $G(y,x)\subseteq G$ is an embedded submanifold. In particular, $G_x$ is a Lie group.
the subset $O_x$ is a (may not be embedded) submanifold in a canonical way.

By a morphism of Lie groupoids $\phi: (\mathcal{G}_1\rightrightarrows \mathcal{G}_0)\rightarrow (\mathcal{H}_1\rightrightarrows \mathcal{H}_0)$ we mean a pair of smooth maps $\phi^{ar}:\mathcal{G}_1\rightarrow \mathcal{H}_1$ and $\phi^{ob}:\mathcal{G}_0\rightarrow \mathcal{H}_0$ compatible with structure maps $s,t,m,u,i$. We write $\phi$ for both $\phi^{ar}$ and $\phi^{ob}$.

What is a Stack?

Sat, 12 Jan 2019 18:18:08 +0000

Given a manifold $M$ we have the concept of open cover of $M$. We usually write an open cover of a manifold $M$ as a collection of open subsets $\{U_i\}$ (such that $\bigcup U_i=M$). In this note we see an open cover of $M$ as a collection of maps (inclusions) $\{U_i\rightarrow M\}$. Some properties of "open cover" are.

(Pull back exists and gives an open cover) Suppose $\{U_i\rightarrow M\}$ is an open cover for $M$ and $\pi:V\rightarrow M$ is a smooth map. Then, $\{\pi^{-1}(U_i) \rightarrow V\}$ is a cover for $V$.
(Diffeomorphisms gives open cover) For any manifold $M$, $M$ itself is considered as an open cover $\{M\rightarrow M\}$. More generally, for any diffeomorphism $M'\rightarrow M$, $\{M'\rightarrow M\}$ is considered as an open cover.
(Open cover of open cover is an open cover) Let $\{U_\alpha\rightarrow U\}$ be an open cover for $U$ i.e., $\bigcup_{\alpha} U_\alpha=U$. Suppose $\{V_{\alpha\beta}\rightarrow U_\alpha\}$ is an open cover for $U_\alpha$ for each $\alpha$ i.e., $\bigcup_{\beta}V_{\alpha\beta}=U_\alpha$. Then, $\bigcup_{\alpha\beta}V_{\alpha\beta}=U$ i.e., $\{V_{\alpha\beta}\rightarrow U\}$ is an open cover for $U$.

For a category $\mathcal{C}$ and an object $U$ of $\mathcal{C}$, a collection of arrows $\{U_i\rightarrow U\}$ is said to be a cover for $U$. Definition : Let $\mathcal{C}$ be a category. A Grothendieck topology on $\mathcal{C}$ is given by a collection of covers $\mathcal{W}=\{\{U_i\rightarrow U\}: U\in \mathcal{C}_0\}$ satisfying following conditions.

(Pullbacks exists and gives a cover) Suppose $\{U_i\rightarrow U\}\in \mathcal{W}$ and $\pi:V\rightarrow U$ be an arrow. Then, the pull back $U_i\times_UV$ exists (as an object in $\mathcal{C}$) and $\{U_i\times_UV \rightarrow V\}$ is a cover for $V$.
(Isomorphisms gives an open cover) Suppose $V\in \mathcal{C}_0$ and $V\rightarrow U$ is an isomorphism in $\mathcal{C}$ then, $\{V\rightarrow U\}\in \mathcal{W}$.
(cover of a cover is a cover) Suppose $\{U_\alpha\rightarrow U\}\in \mathcal{W}$ and $\{U_{\alpha\beta}\rightarrow U_\alpha\}\in \mathcal{W}$ for each $\alpha$. Then, the collection of compositions $\{U_{\alpha\beta}\rightarrow U_\alpha\rightarrow U\}\in \mathcal{W}$.

To talk about a stack over category $\mathcal{C}$ we fix a Grothendieck topology $\mathcal{W}$ on $\mathcal{C}$. When we say cover, we mean it belongs to $\mathcal{W}$. Let $\mathcal{D}$ be a category fibered in groupoids over $\mathcal{C}$ i.e., we have a functor $F:\mathcal{D}\rightarrow \mathcal{C}$ satisfying some conditions.

Given an object $U$ of $\mathcal{C}$ we have what is called fibre of $U$ in $\mathcal{D}$ usually denoted by $\mathcal{D}(U)$.
Given an object $U$ of $\mathcal{C}$ and a cover $\{U_i\rightarrow U\}$ (i.e., it belongs to $\mathcal{W}$) we have what is called descent category associated to the cover $\{U_i\rightarrow U\}$, usually denoted by $\mathcal{D}(\{U_i\rightarrow U\})$.

There is an obvious functor $\mathcal{D}(U)\rightarrow \mathcal{D}(\{U_i\rightarrow U\})$.

Definition : Let $\mathcal{C}$ be a category with Grothendieck topology $\mathcal{W}$. A category fibered in groupoids $\mathcal{D}\rightarrow \mathcal{C}$ is said to be a stack over $\mathcal{C}$ if, for every object $U$ of $\mathcal{C}$ and every cover $\{U_i\rightarrow U\}$, the functor $\mathcal{D}(U)\rightarrow \mathcal{D}(\{U_i\rightarrow U\})$ is an equivalence of categories.

The fibre categroy $\mathcal{D}(U)$ is a category whose objects are that of $\mathcal{D}$ which map to $U$ under $F$ i.e.,

$\mathcal{D}(U)_0=\{V\in \mathcal{D}_0:F(V)=U\}$.

Stacks

Thu, 03 Jan 2019 13:38:28 +0000

Here, I will add links to WordPress pages where I have written something about Stacks. Papers I am reading are

Differentiable Stacks and Gerbes by Kai Behrend and Ping Xu.
Orbifolds as Stacks by Eugene Lerman.
Non abelian Differentiable Gerbes by Camille, Stienon and Ping Xu.
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Criterion for a map of stacks to be an atlas

Thu, 03 Jan 2019 12:51:29 +0000

Definition : A stack $\mathcal{D}\rightarrow \text{Man}$ is differentiable if there exists a manifold $X$ with an atlas $p:\underline{X}\rightarrow \mathcal{D}$ i.e., $p$ is representable surjective submersion.

We see a criterion for a map $p:\underline{X}\rightarrow \mathcal{D}$ to be an atlas. By $p:\underline{X}\rightarrow \mathcal{D}$ to be representable surjective submersion, we mean given a map of stacks $\underline{Y}\rightarrow \mathcal{D}$ the fibered product $\underline{X}\times_{\mathcal{D}}\underline{Y}$ is representable by a manifold and that the map of manifolds $\underline{X}\times_{\mathcal{D}}\underline{Y}\rightarrow \underline{Y}$ is a surjective submersion. As $\underline{X}\times_{\mathcal{D}}\underline{Y}$ is representable by a manifold for any map of stacks $\underline{Y}\rightarrow \mathcal{D}$, in particular, taking $\underline{Y}\rightarrow \mathcal{D}$ to be the same map $\underline{X}\rightarrow \mathcal{D}$ we see that, in particular $\underline{X}\times_{\mathcal{D}}\underline{X}$ is representable by a manifold.

Remark : If $p:\underline{X}\rightarrow \mathcal{D}$ is an atlas for $\mathcal{D}$ then $\underline{X}\times_{\mathcal{D}}\underline{X}$ is representable by a manifold.

As $\underline{X}\times_{\mathcal{D}}\underline{Y}\rightarrow \underline{Y}$ is a submersion for any map of stacks $\underline{Y}\rightarrow \mathcal{D}$, in particular, taking $\underline{Y}\rightarrow \mathcal{D}$ to be the same map $\underline{X}\rightarrow \mathcal{D}$ we see that, projecion map $\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}$ is a submersion. It is not relevant which projection is it as both maps are same. So, both projection maps $pr_1:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}$ and $pr_2:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}$ are submersions.

Remark : If $p:\underline{X}\rightarrow \mathcal{D}$ is an atlas for stack $\mathcal{D}$ then projection maps $pr_1:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}$ and $pr_2:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}$ are submersions.

As any representable surjective submersion is an epimorphism we have following remark.

Remark : If $p:\underline{X}\rightarrow \mathcal{D}$ is an atlas for $\mathcal{D}$ then $p:\underline{X}\rightarrow \mathcal{D}$ is an epimorphism.

Combining all these remarks we have following remark.

If $p:\underline{X}\rightarrow \mathcal{D}$ is an atlas for $\mathcal{D}$ then, $\underline{X}\times_{\mathcal{D}}\underline{X}$ is representable by a manifold and projection maps $pr_1:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}$ and $pr_2:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}$ are submersions and $p:\underline{X}\rightarrow \mathcal{D}$ is an epimorphism.

It turns out that converse of above remark is true.

Proposition : Let $p:\underline{X}\rightarrow \mathcal{D}$ is a morphism of stacks such that $\underline{X}\times_{\mathcal{D}}\underline{X}$ is representable by a manifold and projection maps $pr_1:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}$ and $pr_2:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}$ are submersions and that $p:\underline{X}\rightarrow \mathcal{D}$ is an epimorphism. Then, Then, $p:\underline{X}\rightarrow \mathcal{D}$ is a representable surjective submersion i.e., an atlas for $\mathcal{D}$.

Before we give proof of this, we recall a result.

Lemma : Let $\mathcal{D}\rightarrow\mathcal{C}$ be a morphism of stacks. Suppose $U$ be a manifold and $\underline{U}\rightarrow \mathcal{C}$ is an epimorphism of stacks such that fiber product $\mathcal{D}\times_{\mathcal{C}}\underline{U}$ is represented by a manifold and the map of manifolds $\mathcal{D}\times_{\mathcal{C}}\underline{U}\rightarrow U$ is a submersion. Then, $\mathcal{D}\rightarrow \mathcal{C}$ is a representable submersion.

To prove $\underline{X}\rightarrow \mathcal{D}$ is a representable submersion, consider an epimorphism of stacks, namely $\underline{X}\rightarrow \mathcal{D}$ (it is given to be an epimorphism, condition $2$ above). See that the fibre product $\underline{X}\times_{\mathcal{D}}\underline{X}$ is representable by a manifold (it is given in condition $1$ above) and that the projection map $\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow X$ is a submersion (it is in condition $1$ above). Thus, by above lemma, $p:\underline{X}\rightarrow \mathcal{D}$ is a representable submersion. Note that, a representable submersion that is an epimorphism is a representable surjective submersion. Thus, $p:\underline{X}\rightarrow \mathcal{D}$ is an atlas for $\mathcal{D}$. So, we have the following result.

Proposition : Let $p:\underline{X}\rightarrow \mathcal{D}$ is a morphism of stacks such that $\underline{X}\times_{\mathcal{D}}\underline{X}$ is representable by a manifold and projection maps $pr_1:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}$ and $pr_2:\underline{X}\times_{\mathcal{D}}\underline{X}\rightarrow \underline{X}$ are submersions and that $p:\underline{X}\rightarrow \mathcal{D}$ is an epimorphism. Then, Then, $p:\underline{X}\rightarrow \mathcal{D}$ is a representable surjective submersion i.e., an atlas for $\mathcal{D}$.

Criterion for a stack to be representable

Wed, 02 Jan 2019 15:20:43 +0000

Criterion for a map to be representable submersion

Mon, 31 Dec 2018 17:53:25 +0000

A morphism of stacks $f:\mathcal{D}\rightarrow \mathcal{C}$ is called a representable submersion if, for every morphism $\underline{M}\rightarrow \mathcal{C}$, the fibred product $\mathcal{D}\times_{\mathcal{C}}\underline{M}$ is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds $\mathcal{D}\times_{\mathcal{C}}\underline{M}\rightarrow M$ is a submersion. Following is a criterion for a map of stacks to be representable submersion. The result says it is enough to check for epimorphisms $\underline{M}\rightarrow \mathcal{C}$. Precise statement is as follows.

Let $f:\mathcal{D}\rightarrow \mathcal{C}$ be a morphism of stacks. Suppose given a manifold $U$ and a morphism of stacks $\underline{U}\rightarrow \mathcal{C}$ which is an epimorphism. If the fibered product $\mathcal{D}\times_{\mathcal{C}}\underline{U}$ is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds $\mathcal{D}\times_{\mathcal{C}}\underline{U}\rightarrow U$ is a submersion, then $f$ is representable submersion.

Let us see what this means in the set up of manifolds. Let $f:\mathcal{D}\rightarrow \mathcal{C}$ be a morphism of stacks. Let $f:M\rightarrow N$ be a morphism of manifolds (which gives a morphism of stacks $\underline{M}\rightarrow \underline{N}$). Suppose given a manifold $U$ and a morphism of stacks $\underline{U}\rightarrow \mathcal{C}$ which is an epimorphism. A representable surjective submersion is an epimorphism. So, we consider a surjective submersion $g:U\rightarrow N$ (which gives an epimorphism $\underline{U}\rightarrow \underline{N}$ being a representable surjective submersion). If the fibered product $\mathcal{D}\times_{\mathcal{C}}\underline{U}$ is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds $\mathcal{D}\times_{\mathcal{C}}\underline{U}\rightarrow U$ is a submersion. As $U\rightarrow N$ is submersion, it is anyways true that $M\times_N U$ is a smooth manifold. What is extra that we have here is that $M\times_NU\rightarrow U$ is a submersion. It is anyways true that $M\times_NU\rightarrow M$ is a submersion being a pullback of submersion. But it is not true in general that $M\times_NU\rightarrow U$ is a submersion. Here, we are given that $M\times_NU\rightarrow U$ is a submersion. So,

Let $f:\mathcal{D}\rightarrow \mathcal{C}$ be a morphism of stacks. Suppose given a manifold $U$ and a morphism of stacks $\underline{U}\rightarrow \mathcal{C}$ which is an epimorphism. If the fibered product $\mathcal{D}\times_{\mathcal{C}}\underline{U}$ is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds $\mathcal{D}\times_{\mathcal{C}}\underline{U}\rightarrow U$ is a submersion, then $f$ is representable submersion.

turns to

Let $f:M\rightarrow N$ be a morphism of manifolds be a morphism of manifolds. Suppose given a manifold $U$ and a sujective submersion $\underline{U}\rightarrow N$. If the fibered product $M\times_{N}U$ is a manifold and that the map of manifolds $M\times_{N}U\rightarrow U$ is a submersion, then $f$ is submersion.

This is more or less obvious. We have following commutative diagram As $G:U\rightarrow N$ is a submersion (we have started with this) and $p_2: M\times_N U\rightarrow U$ is a submersion (we are given this), the composition $G\circ p_2=F\circ p_1$ is a submersion which then imply that $F:M\rightarrow N$ is a submersion. Let $m\in M$. As $G:U\rightarrow N$ is surjective, so is $p_1$ (pullback of surjective is surjective) i.e., there exists $(m,u)\in M\times_N U$ such that $p_1(m,u)=m$. As $F\circ p_1$ is submersion, $(F\circ p_1)_{*,(m,u)}(T_{m,u}(M\times_N U))=T_{F(m)}N$. Applying chain rule, we have $F_{*,m}((p_1)_{*,(m,u)}(T_{m,u}(M\times_N U)))=T_{F(m)}N$, in particular, $F_{*,m}(T_mM)=T_{F(m)}N$. Thus, $F$ is submersion. So, we need both surjectivity and submersion of $U\rightarrow N$. Now, let us look at more general case. Now, $U\rightarrow N$ is not a surjective submersion but induces an epimorphism $U\rightarrow N$. Suppose that the pullback $M\times_N U$ is a manifold and that the map $M\times_N U\rightarrow U$ is a submerson. Let $W\rightarrow N$ be a map. We need to prove that $M\times_N W$ is a manifold. We have following diagram As $M\times_N U\rightarrow U$ is a submersion, the pullback $(M\times_N U)\times_U W_i=M\times_N W_i$ is a manifold. So, we have an open cover $\{W_i\rightarrow W\}$ of $W$ such that the pullbacks $M\times_N W_i$ are manifolds. I think this should confirm that $M\times_N W$ is a manifold and just because $M\times_N W\rightarrow W_i$ are submersions, so is the map $M\times_N W\rightarrow W$. Thus, $f:M\rightarrow N$ is a representable submersion. The same idea works for an arbitrary map of stacks $\mathcal{D}\rightarrow \mathcal{C}$. Let $\underline{W}\rightarrow \mathcal{C}$ be a map of stacks. We have to prove that $\mathcal{D}\times_{\mathcal{C}}\underline{W}$ is representable and that the map of manifolds $\mathcal{D}\times_{\mathcal{W}}\underline{W}\rightarrow W$ is a submersion. As $\underline{U}\rightarrow \mathcal{C}$ is epimorphism, for $\underline{W}\rightarrow \mathcal{C}$ there exists an open cover $\{W_i\rightarrow W\}$ with commutative diagram as shown below. We have following diagram As $\mathcal{D}\times_{\mathcal{C}} \underline{U}\rightarrow U$ is a submersion, the pullback $(\mathcal{D}\times_{\mathcal{C}} \underline{U})\times_U W_i=\mathcal{D}\times_{\mathcal{C}} W_i$ is a manifold. So, we have an open cover $\{W_i\rightarrow W\}$ of $ W$ such that the

$(\mathcal{D}\times_{\mathcal{C}}W)\times_W W_i=\mathcal{D}\times_{\mathcal{C}} W_i$

Math stack exchange/ stack overflow questions

Tue, 11 Jul 2017 14:33:19 +0000

Here I will add links of pages of interesting questions/answers from Math Stack Exchange and Math over flow.