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