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Lecture note on classification of covering space

covering map covering space classification

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#1 bangbang1412

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Đã gửi 24-02-2017 - 16:28

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

Declare to yourself that, from now on, your life is dedicated to one and only one woman, the greatest mistress of your life, the tenderest woman you have ever encountered, Mathematica.


#2 Isidia

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Đã gửi 04-09-2017 - 13:31

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

Chính là gì, là đúng đắn. Nghịch là gì, là sai lệch. Đạo là gì, là đường đi. Chính đạo là đường đi đúng đắn dẫn người ta đến với lẽ phải. Nghịch đạo là lối mòn sai lệch dẫn người ta vào chốn lầm lạc. Đạo của việc học là để thị biệt đúng sai, phân định đâu là chính đạo, đâu là nghịch đạo. Than ôi, đường đời rối như trăm mối tơ vò, chính đạo chỉ có một mà nghịch đạo có cả trăm, biết theo lối nào đây? Cho nên, đạo của bậc làm thầy là dẫn lối cho kẻ học trò đến gần với chính đạo, can ngăn kẻ ngu ngơ rời xa khỏi nghịch đạo. Hướng về chính đạo, rời xa nghịch đạo há chỉ dừng ở đường thi thôi, mà còn phải dẫn lối suy nghĩ. Ngẫm sâu rồi mới nghe thầy, nghĩ sâu rồi mới theo thầy, đó là chính đạo của việc học . Chưa hiểu mà đã dạ vâng, tồn nghi mà không chịu hỏi, đó chính là nghịch đạo của phận học trò.





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