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# Mobius band

The green arrows in the image above indicate how the two edges are attached to each other. If you reverse the orientation, you will get the Mobius band ${\bf M}$ (aka "Mobius strip", or, rather "Möbius").

The equivalence relation is given by

$(x,y) \sim (u,v)$ if $y=1-v$ and $x,u=0$ or $1$, or simply

$(0,y) \sim (1,1-y).$

The Mobius band has one side! If you draw a line with a marker along the center of the band, you'll arrive to the spot where you started, just like with the cylinder, but "both sides" will be marked as a result.

Theorem. The Mobuis band is a surface with boundary.

What happens if we cut the Mobius band along this line? The result is a thinner band with a double twist.

Since this band can be cut, twisted $360$ degrees, and then glued back together along the same edge - exactly as before! - this double twist band is homeomorphic to the cylinder.

Another way to see this is shown in the diagram below:

Exercise. What happens if we cut the Mobius band by two, three, $n$ cuts along the middle line? Solution: Cutting the Mobus band.

Technically, the Mobius band is a $2$-dimensional manifold, i.e., surface, with boundary. The boundary is homeomorphic to the circle.

Let's represent it as a cell complex. We can interpret the gluing as equivalence of $a$ and $c$. But this time they are attached to each other with $c$ upside down. It makes sense then to interpret this as an equivalence:

$$a \sim -c,$$

$$A \sim D, B \sim C.$$

The boundary operator is:

$$\partial \tau = a + b + c + d = a + b - a + d = b + d,$$

$$\partial a = A + B, \partial b = B + C = 0, \partial c = A + B, \partial d = D + A = 0$$

Then the homology of the Mobius band is:

$$H_1({\bf M})= {\bf Z}.$$

Same as the circle! Why is that? You can see the reason why by realizing that we can contract the band to the middle line without changing its homology (see Homotopy equivalence).

It is obvious that the middle line is homeomorphic to the circle. Its boundary however appears to be made of two pieces, just like that of the cylinder.

Theorem. The boundary of the Mobius band is homeomorphic to the circle.

Proof. You travel along the "lower edge" from $A$ to $B$, and next you are at $B$ on the "upper" edge, the you travel along it to $A$. You come back where you started. QED

Exercise. Compute the homology of Möbius band.

Exercise. Compute the homology of Möbius band relative its edge.

Exercise. Show that the middle line is a deformation retract of the Möbius band.

Exercise. Compute the homology map of the composition of the inclusion of the boundary of the Möbius band and its retraction to the middle line.