Đến nội dung

Hình ảnh

Lecture note on classification of covering space

- - - - - covering map covering space classification

  • Please log in to reply
Chủ đề này có 1 trả lời

#1
bangbang1412

bangbang1412

    Độc cô cầu bại

  • Phó Quản lý Toán Cao cấp
  • 1667 Bài viết

Learn more mathematics I seem realize the way they quest is very direct , natural ( including soluntion , how to solve problem ....) . I note here about all of covering space and covering map 

Definition : a continuous serjective map $p : E \to B$ . The open set $U$ of $B$ is said can be evenly covered by $p$ if the iverse image $p^{-1}(U)$ can be break into disjoint sets which each of them is homeomorphic to $U$ map by $p$ . The collection of sets is called a partion into slice of $U$ . It is called covering space ( map ) if each point $b \in B$ has a neighborhood that can be evenly covered by $p$ . 

A nature of covering map is " it's a open map " which means it maps open set to open set .

The most importance covering map is the map : 

$$p : R \to S^{1}$$

$$p(x) = (cos 2\pi x, sin 2\pi x)$$

It is called fundamental covering map , which is useful in calculus the fundamental group of $S^{1}$ ( $\pi_{1}(S^{1},x_{0}) \cong Z$ )

Definition : Given two maps :

$$ p : E \to B$$

$$f  : X \to B$$

A lifting of the map $f$ is a map :

$$\overline{f} : X \to E$$

such that $p.\overline{f} = f$

Lemma $1$ : A covering map $p : (E , e_{0}) \to ( B , b_{0})$ then every paths $f : [0,1] \to B$ beginning at $b_{0}$ has a unique lifting to a path $\overline{f}$ in E beginning at $e_{0}$ 

Lemma $2$ : A covering map $p : ( E,e_{0}) \to (B , b_{0})$ , let the map $F : [0,1] \times [0,1] \to B$ be continuous such that $F(0,0)=b_{0}$ also has a unique lifting to a map $\overline{F}$ in $E$ 

$$\overline{F} : [0,1] \times [0,1] \to E$$

such that $\overline{F}(0,0) = e_{0}$ . If $F$ is a path homotopy then $\overline{F}$ too . 

Lemma $3$ : Let $p : (E,e_{0})  \to (B,b_{0})$ be a covering map . Let two path $f,g$ in $B$ from $b_{0}$ to $b_{1}$ , let $\overline{f},\overline{g}$ be their respective liftings to path in $E$ begin at $e_{0}$ . If $f \sim g$ ( in an path equivalent homotopy ) then $\overline{f},\overline{g}$ end at the same point in $E$ and also path-homotopic .

From this lemma , it introduces us to a new definition , a map which can used to research fundamental group . Given an element $[f]$ in $\pi_{1}(B,b_{0})$ then the lifting correspondence is well-defined as follow : 

$$\phi : \pi_{1}(B,b_{0}) \to p^{-1}(b_{0})$$

$$\phi([f]) \to \overline{f}(1)$$

If $E$ is path-connected then $\phi$ is surjective , if $E$ is simply connected , it's a bijection . So it's easy to see the fundamental $\pi_{1}(S^{1}) \cong Z$ . The following theorem will introduce us to classification theorem of covering space . The first , we use covering map to research fundamental group then we use ( opposite ) fundamental group to research covering space , the purpose is transform topology to algebra which is easier . 

Theorem $1$ : Let $p : (E,e_{0}) \to (B,b_{0})$ be a covering map then 

$a)$ The map is introduced by $p$ : 

$$p_{*} : \pi_{1}(E,e_{0}) \to \pi_{1}(B,b_{0})$$

$$p_{*}([f]) = [p.f]$$

is an monomorphism 

$b)$ $H=p_{*}(\pi_{1}(E,e_{0})) \subset B$ then the lifting correspondence induces and injective map ( which bijective if $E$ is simply connected ) 

$$\Phi : \pi_{1}(B,b_{0}) / H \to p^{-1}(b_{0})$$

$c)$ A loop $f$ in B based at $b_{0}$ , then $[f] \in H$ if and only if $f$ lifts to a loop in $E$ based at $e_{0}$

From now we have a convetion that $E,B$ is path-connected and locally connected ( fortunely we usually research on spaces like that , you will see how it's important when you try to prove following theorems ) .

Theorem $2$ : Let $p : (E,e_{0}) \to (B,b_{0})$ be a covering map . Let $f : (Y,y_{0}) \to (B,b_{0})$ be a continuous map. Suppose that $Y$ is also locally path-connected and path-connected . The map $f$ can be lift to a map $\overline{f} : (Y,y_{0}) \to (E,e_{0})$ if and only if :

$$f_{*}(\pi_{1}(Y,y_{0})) \subset p_{*}(\pi_{1}(E,e_{0}))$$

Two covering map $p,p'$ from $E,E'$ to $B$ said to be equivalent if and only if exist a homeomorphism $h$ from $E$ to $E'$ such that $p = p'.h$

Theorem $3$ : Let two covering maps : 

$$p : (E,e_{0}) \to (B,b_{0})$$

$$p' : (E,e'_{0}) \to (B,b_{0})$$

There is a homeomorphic equivalence such that it maps $e_{0}$ to $e'_{0}$ if and only if ( if exist , it's unique ) 

$$\begin{equation}p_{*}(\pi_{1}(E,e_{0})) = p'_{*}(\pi_{1}(E',e'_{0})) \end{equation}$$

Then from this theorem and a bit lemma we can show $h$ exists ( don't need the condition $h(e_{0})=e'_{0}$) if and only if two subgroups at $(1)$ is conjugate . 


Bài viết đã được chỉnh sửa nội dung bởi bangbang1412: 25-02-2017 - 21:52

$$[\Psi_f(\mathbb{1}_{X_{\eta}}) ] = \sum_{\varnothing \neq J} (-1)^{\left|J \right|-1} [\mathrm{M}_{X_{\sigma},c}^{\vee}(\widetilde{D}_J^{\circ} \times_k \mathbf{G}_{m,k}^{\left|J \right|-1})] \in K_0(\mathbf{SH}_{\mathfrak{M},ct}(X_{\sigma})).$$


#2
Isidia

Isidia

    Hạ sĩ

  • Thành viên
  • 74 Bài viết
Learn more mathematics I seem realize the way they quest is very direct , natural ( including soluntion , how to solve problem ....) . I note here about all of covering space and covering map

 

Grammatical correction of your first sentence:

The more I learn mathematics, the more it becomes apparent to me that the quest is very direct and natural ( for solution , how to solve problems etc). What I note down here is about covering space and covering map.

 

Well, your English is not that bad. I can see that you are able to absorb fragments of English syntactical structures from reading numerous mathematical research papers. But let's just say that you are not going to be able to write a research paper based on this kind of broken English. Sometimes I just wonder if this is written by an autistic person. But hey, if autism's side effect is incredible mathematical talent, then more power to you.

 

But, you know, some guys out there can be both a maestro in Mathematics, and become fluent in several foreign languages. Think of Andrey Kolmogorov for instance.


Bài viết đã được chỉnh sửa nội dung bởi Isidia: 04-09-2017 - 13:39

There is no mathematical model that can predict your future or tell you how your life will unfold. All strength and power lies within your soul, and that's all what you need.





2 người đang xem chủ đề

0 thành viên, 2 khách, 0 thành viên ẩn danh