Re: [分析] 一個高微與圖形的問題
※ 引述《znmkhxrw (QQ)》之銘言：
: 我想要證明(或找到反例)以下敘述：
: Let D = {(x,y)│x^2+y^2<=1} , S = {(x,y)│x^2+y^2=1}
: and A,B be disjoint compact subsets of D with A∩S = {(0,1)}, B∩S = {(0,-1)}
: Show that every point P in left arc is path-connected in D
: to every point Q in right arc without passing through A∪B.
: (that is, there exists continuous f：[0,1]→ D\(A∪B) with f(0)=P, f(1)=Q )
: <note>
一位網友 Dejan Govc 造的函數
f:D->S 應該說 f 導出的基本群的homomorphism 不是nontrivial
符號
D={(x,y)|x^2+y^2<= 1} , S={(x,y)│x^2+y^2=1}
T1 = left arc = {exp(i*x)│ 0.5pi<x<1.5pi}
T2 = right arc= {exp(i*x)│-0.5pi<x<0.5pi}
T3 = {exp(i*x)│0<x<pi}
T4 = {exp(i*x)│pi<x<2pi}
A,B : disjoint compact subsets of D
a = A∩S = {(0,1)}, b = B∩S= {(0,-1)}
──
C1:the component of D\(A∪B) containing T1 .
"Suppose T2 is not contained in C1"
Let E = boundary(C1)
then E is contained in A∪B
Let A1= A ∩ E
B1= B ∩ E
Define f:D->S as follows
1 d(x,A1)
g(x)= ── ───────── d:|R^2 usual metric
2 d(x,A1)+d(x,B1)
＿
| exp( i*2p:i*g(x) ) x:Closure(C1) = C1
f(x)=| ＿＿
| exp( i*2pi*(1-g(x)) ) x:Closure(D\C1)= D\C1
then f is a "well-defined continuous" function f:D->S
f(a)=1
f(b)=-1 ＿ ＿ ＿
C1 connected =>f(C1) = T3
＿ ＿＿ ＿＿ ＿
T2 is contained in D\C1 =>f(D\C1) = T4
Consider this fundamental group h:[0,1]->S , h(x)=exp(i*2pi*x - 0.5pi )
f*:[h]-> [f(h)]
Using lifting lemma , it is easy to know that
f(h) is not contractible
Hence f*: π(D)->π(S) is not trivial.
a contradiction !
so C1 must contain T2
Note that C1\S is an open set
and in this case C1\S is also connected
In |R^n , open connected => path connected
so C1\S path connected => C1 path connected
＃
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※ 編輯: keroro321 (1.163.138.85), 02/02/2016 18:49:25
※ 編輯: keroro321 (1.163.138.85), 02/02/2016 19:10:40
※ 編輯: keroro321 (1.163.138.85), 02/02/2016 20:01:44
※ 編輯: keroro321 (1.163.138.85), 02/02/2016 20:08:53
※ 編輯: keroro321 (1.163.138.85), 02/02/2016 20:16:56
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