Klein bottle contains Möbius band

hbghlyj

A simple closed curve $C$ in a space $X$ is the image of a continuous injection $S^1→X$. Find simple closed curves $C_1, C_2$ and $C_3$ in the Klein bottle $K$ such that

• $K∖C_1$ has one component, which is homeomorphic to an open annulus $S^1×(0,1)$.
• $K∖C_2$ has one component, which is homeomorphic to an open Möbius band.
• $K∖C_3$ has two components, each of which is homeomorphic to an open Möbius band.

[An open Möbius band is the space obtained from $[0,1] \times(0,1)$ by identifying $(0, y)$ with $(1,1-y)$ for each $y \in(0,1)$.]

hbghlyj

If Klein bottle is cut along the dashed curve $C$
(it is closed curve because 4 corners are identified)
size(4cm);
pair A = (0,0);
pair B = (1,0);
pair C = (1,1);
pair D = (0,1);
currentpen=blue;
fill(box(A,C),lightgray);
draw(A--B,MidArrow);
draw(C--D,MidArrow);
currentpen=red;
draw(C--B,MidArrow);
draw(D--A,MidArrow);
draw(arc((1,0.5),0.5,90,270),dashed);

$K\setminus C$ has 1 component:size(4cm);
pair A = (0,0);
pair B = (1,0);
pair C = (1,1);
pair D = (0,1);
path bigarc=arc((1,0.5),0.5,90,270);
path smallarc=shift(1,0)*xscale(.5)*arc((0,0.5),0.5,90,270);
fill(A--reverse(bigarc)--smallarc--C--D--cycle,evenodd+lightgray);
currentpen=blue;
draw(A--B,MidArrow);
draw(C--D,MidArrow);
currentpen=red;
draw(C--B,MidArrow);
draw(D--A,MidArrow);
draw(bigarc);
draw(smallarc);
We can shift the right part to the left (as points are identified, this gives the same space):
size(4cm);
pair A = (0,0);
pair B = (1,0);
pair C = (1,1);
pair D = (0,1);
path bigarc=arc((1,0.5),0.5,90,270);
path smallarc=xscale(.5)*arc((0,0.5),0.5,90,270);
fill(A--reverse(bigarc)--C--smallarc--cycle,evenodd+lightgray);
currentpen=blue;
draw(A--B,MidArrow);
draw(C--D,MidArrow);
currentpen=red;
draw(bigarc);
draw(smallarc);
This is a Möbius band. So part (b) is solved.

hbghlyj

If Klein bottle is cut along the dashed curve $C$
(it is closed curve because 4 corners are identified)
size(4cm);
pair A = (0,0);
pair B = (1,0);
pair C = (1,1);
pair D = (0,1);
currentpen=blue;
fill(box(A,C),lightgray);
draw(A--B,MidArrow);
draw(C--D,MidArrow);
currentpen=red;
draw(C--B,MidArrow);
draw(D--A,MidArrow);
draw(arc((0.5,1),0.5,180,360),dashed);

$K\setminus C$ has 1 component: size(4cm);
pair A = (0,0);
pair B = (1,0);
pair C = (1,1);
pair D = (0,1);
path bigarc=arc((0.5,1),0.5,180,360);
path smallarc=shift(0,1)*yscale(.5)*arc((0.5,0),0.5,180,360);
fill(A--B--reverse(bigarc)--smallarc--D--cycle,evenodd+lightgray);
currentpen=blue;
draw(A--B,MidArrow);
draw(C--D,MidArrow);
currentpen=red;
draw(C--B,MidArrow);
draw(D--A,MidArrow);
draw(bigarc);
draw(smallarc);size(4cm);pair A = (0,0);pair B = (1,0);pair C = (1,1);pair D = (0,1);path arc=arc((0.5,1),0.5,180,360);fill(arc--B--A--cycle,evenodd+lightgray);draw(A--B,blue,MidArrow);draw(C--B,red,MidArrow);draw(D--A,red,MidArrow);draw(arc);path upper=shift(0,.2)*arc;fill(upper--cycle,evenodd+lightgray);draw(upper);draw(shift(0,.2)*(C--D),blue,MidArrow);
We can rotate the upper part 180° and stick to the lower (as points are identified, this gives the same space):
size(4cm);pair A = (0,0);pair B = (1,0);pair C = (1,1);pair D = (0,1);path upper=arc((0.5,1),0.5,180,360);path lower=shift(0,-1)*upper;fill(reverse(lower)--upper--cycle,evenodd+lightgray);draw(C--B,red,MidArrow);draw(D--A,red,MidArrow);draw(upper);draw(lower);
This is homeomorphic to an open annulus $S^1×(0,1)$.

hbghlyj

math.stackexchange.com/questions/1039875
can cut the Klein Bottle along the red edge (in reverse, a Möbius band becomes Klein Bottle when the two red edges are identified).

part a) second solution

can cut the Klein Bottle along the blue edge (in reverse, a cylinder becomes Klein Bottle when the two blue edges are identified).

Czhang271828

Here is a short-cut to understand why Klein bottle contains Möbius band:

(1) Klein bottle = Möbius band + quotient relation on its boundary,

(2) There is another Möbius band properly contained in a given Möbius band.

Therefore, Klein bottle contains Möbius band.

hbghlyj

Cutting vertically in the middle

size(4cm);
pair A = (0,0);
pair B = (1,0);
pair C = (1,1);
pair D = (0,1);
currentpen=blue;
fill(box(A,C),lightgray);
draw(A--B,MidArrow);
draw(C--D,MidArrow);
currentpen=red;
draw(C--B,MidArrow);
draw(D--A,MidArrow);
draw((0.5,1)--(0.5,0),dashed);

we get 2 Möbius bands
unitsize(4cm);
picture pic;unitsize(pic,4cm);
pair A = (0,0);
pair B = (1/2,0);
pair C = (1/2,1);
pair D = (0,1);
fill(pic,box(A,C),lightgray);
draw(pic,A--B,blue,MidArrow);
draw(pic,C--D,blue,MidArrow);
picture pic2;unitsize(pic2,4cm);
add(pic2,pic.fit(),(1/4,1/2),(0,0));
draw(pic,D--A,red,MidArrow);
draw(pic2,C--B,red,MidArrow);
add(pic.fit(),(0,0),E);
add(pic2.fit(),(1,0),E);

hbghlyj

$C$ is the dashed curve (it is closed curve because 4 corners are identified, and midpoints of opposite sides are identified)
size(4cm);pair A = (0,0);pair B = (1,0);pair C = (1,1);pair D = (0,1);fill(box(A,C),lightgray);draw(A--B,blue,MidArrow);draw(C--D,blue,MidArrow);draw(C--B,red,MidArrow);draw(D--A,red,MidArrow);draw(B/2--D,dashed);draw(B/2+D--B,dashed); $K∖C$ is the shaded region: size(0,4cm);pair A = (0,0);pair B = (1/2,0);pair C = (1/2,1);pair D = (0,1);fill(A--B--D--cycle,lightgray);fill(shift(.2)*(C--2B--B--D--cycle),lightgray);fill(shift(.9,0)*(B--C--D--cycle),lightgray);draw(shift(.2)*(B--2B),blue,MidArrow);draw(shift(.2)*(C--2B));draw(shift(.2)*(D--B));draw(shift(.2)*(C--D),blue,MidArrow);draw(A--B,blue,MidArrow);draw(shift(.9,0)*(C--D),blue,MidArrow);draw(shift(.9,0)*(C--B),red,MidArrow);draw(D--A,red,MidArrow);draw(B--D);draw(shift(.9,0)*(B--D));
$K∖C$ has only 2 components because the right edge is glued to the left edge:
size(0,4cm);pair A = (0,0);pair B = (1/2,0);pair C = (1/2,1);pair D = (0,1);fill(A--B--D--D-B--cycle,lightgray);fill(shift(.2)*(C--2B--B--D--cycle),lightgray);draw(shift(.2)*(B--2B),blue,MidArrow);draw(shift(.2)*(C--2B));draw(shift(.2)*(D--B));draw(shift(.2)*(C--D),blue,MidArrow);draw(A--B,blue,MidArrow);draw(shift(-.5,0)*(C--D),blue,MidArrow);draw(B--D);draw(shift(-.5)*(B--D)); Each is homeomorphic to a Möbius band.