(1):2 ∣ n 2|n 2∣ n S 1 × I S^1\times I S 1 × I 2 ∤ I 2\not |I 2 ∣ I
(2):打结不改变其同胚性.还是圆环面或莫比乌斯带.
长条形左右两端都是I = [ 0 , 1 ] I=[0,1] I = [ 0 , 1 ] I I I
设f : I → I f:I\to I f : I → I g : I → I , g = ( x ↦ x ) g:I\to I,g=(x\mapsto x) g : I → I , g = ( x ↦ x ) I I I f ∘ g f\circ g f ∘ g I I I I I I I I I g g g g − 1 g^{-1} g − 1 I × I / ( ( 0 , x ) ∼ ( 1 , x ) ) I\times I/((0,x)\sim (1,x)) I × I / (( 0 , x ) ∼ ( 1 , x )) I × I / ( ( 0 , x ) ∼ ( 0 , 1 − x ) ) I\times I/((0,x)\sim (0,1-x)) I × I / (( 0 , x ) ∼ ( 0 , 1 − x ))
考虑莫比乌斯带同胚于I × I / ∼ I\times I/\sim I × I / ∼ x ∼ y ⟺ x = y ∨ ( ∃ t s . t . { x , y } = { ( 0 , t ) , ( 1 , 1 − t ) } ) x \sim y \iff x=y\lor (\exists t \ s.t.\ \{ x,y \} =\{ (0,t),(1,1-t) \} ) x ∼ y ⟺ x = y ∨ ( ∃ t s . t . { x , y } = {( 0 , t ) , ( 1 , 1 − t )})
那么莫比乌斯带的二等分腰线是一条( 0 , 1 2 ) ⇝ ( 1 , 1 2 ) (0,\dfrac12)\rightsquigarrow (1,\dfrac12) ( 0 , 2 1 ) ⇝ ( 1 , 2 1 ) α ( t ) = ( t , 1 2 ) \alpha(t)=(t,\dfrac12) α ( t ) = ( t , 2 1 )
于是剪开后的莫比乌斯带即:
( [ 0 , 1 2 ) ∪ ( 1 2 , 1 ] ) × I / ∼ \begin{gathered}
([0,\dfrac12)\cup (\dfrac12,1])\times I/\sim
\end{gathered} ([ 0 , 2 1 ) ∪ ( 2 1 , 1 ]) × I / ∼ 于是可构造同胚:
h : ( [ 0 , 1 2 ) ∪ ( 1 2 , 1 ] ) × I / ∼ → S 1 × I h ( S ) = { ( e π x i , 2 y ) , if S = { ( x , y ) } , x ∈ ( 0 , 1 ) ∧ y < 1 2 ( − e π x i , 2 − 2 y ) , if S = { ( x , y ) } , x ∈ ( 0 , 1 ) ∧ y > 1 2 ( 1 , 2 y ) , if S = { ( 0 , y ) , ( 1 , 1 − y ) } , y < 1 2 ( − 1 , 2 − 2 y ) , if S = { ( 0 , y ) , ( 1 , 1 − y ) } , y > 1 2 \begin{gathered}
h:([0,\dfrac12)\cup (\dfrac12,1])\times I/\sim\to S^1\times I \\
h(S)=\begin{cases}
(e^{\pi x i},2y) ,\text{ if } S=\{(x,y)\},x\in (0,1)\land y<\dfrac12 \\
(-e^{\pi x i},2-2y) ,\text{ if } S=\{(x,y)\},x\in (0,1)\land y>\dfrac12 \\
(1,2y) ,\text{ if } S=\{(0,y),(1,1-y)\},y<\dfrac12 \\
(-1,2-2y) ,\text{ if } S=\{(0,y),(1,1-y)\},y>\dfrac12
\end{cases}
\end{gathered} h : ([ 0 , 2 1 ) ∪ ( 2 1 , 1 ]) × I / ∼→ S 1 × I h ( S ) = ⎩ ⎨ ⎧ ( e π x i , 2 y ) , if S = {( x , y )} , x ∈ ( 0 , 1 ) ∧ y < 2 1 ( − e π x i , 2 − 2 y ) , if S = {( x , y )} , x ∈ ( 0 , 1 ) ∧ y > 2 1 ( 1 , 2 y ) , if S = {( 0 , y ) , ( 1 , 1 − y )} , y < 2 1 ( − 1 , 2 − 2 y ) , if S = {( 0 , y ) , ( 1 , 1 − y )} , y > 2 1 容易验证它随每个分量连续,且是双射,且逆映射连续.
克莱因瓶即I × I / ∼ , x ∼ y ⟺ ∃ t , { x , y } = { ( 0 , t ) , ( 1 , t ) } ∨ { x , y } = { ( t , 0 ) , ( 1 − t , 1 ) } I\times I/\sim,x\sim y \iff \exists t,\{ x,y \} =\{ (0,t),(1,t) \} \lor \{ x,y \} =\{ (t,0),(1-t,1) \} I × I / ∼ , x ∼ y ⟺ ∃ t , { x , y } = {( 0 , t ) , ( 1 , t )} ∨ { x , y } = {( t , 0 ) , ( 1 − t , 1 )}
那么直接定义映射T : I × I / ∼ ⟹ im f , T ( S ) = f ( s ) , s ∈ S T:I\times I/\sim \implies \operatorname{im} f,T(S)=f(s),s\in S T : I × I / ∼ ⟹ im f , T ( S ) = f ( s ) , s ∈ S
证明T T T f ( 0 , t ) = ( 1 , cos ( 2 π t ) , sin ( 2 π t ) , 0 , 0 ) = f ( 1 , t ) f(0,t)=(1,\cos(2\pi t),\sin(2\pi t),0,0)=f(1,t) f ( 0 , t ) = ( 1 , cos ( 2 π t ) , sin ( 2 π t ) , 0 , 0 ) = f ( 1 , t ) f ( t , 0 ) = ( cos ( 2 π t ) , 1 , 0 , sin ( 2 π t ) , 0 ) = f ( 1 − t , 1 ) f(t,0)=(\cos(2\pi t),1,0,\sin(2\pi t),0)=f(1-t,1) f ( t , 0 ) = ( cos ( 2 π t ) , 1 , 0 , sin ( 2 π t ) , 0 ) = f ( 1 − t , 1 )
仍然观察到T T T T T T
T T T ∀ f ( x , y ) ∈ im f , T π ( x , y ) = f ( x , y ) \forall f(x,y)\in \operatorname{im} f,T\pi(x,y)=f(x,y) ∀ f ( x , y ) ∈ im f , T π ( x , y ) = f ( x , y )
T T T T ( a , b ) = T ( x , y ) T(a,b)=T(x,y) T ( a , b ) = T ( x , y ) b = y b=y b = y x = a x=a x = a
T T T T T T
得到一个莫比乌斯环和一个平环.