Why are the Möbius strip and the boundary of a Klein bottle homotopy equivalent to $S^1$

Question

I'm trying to calculate the fundamental group of two Möbius strips which have been identified along their boundary (which is a Klein bottle, I think). I've chosen an NDR pair $A,B$ where $A$ and $B$ are each of the original Möbius strips. The intersection $A \cap B$ is the boundary of the Klein bottle. I've heard that this boundary, and each of the Möbius strips, are contractible to $S^1$, but I'm having trouble visualizing this. Any help would be appreciated. Thanks

Answer 1

Imagine the Möbius strip as the unit square, where $(0,x)$ is identified with $(1,1-x)$. The line $\{(x,\frac{1}{2}):x\in[0,1]\}$ is a circle as a subspace of the Möbius strip. Then the map $H:M\times[0,1]\to M$ given by $H((x,y),t)=(x,(1-t)y+\frac{t}{2})$ gives a strong deformation retract of $M$ to the circle.

Below is a picture of the Klein bottle cut into two Möbius strips (along the orange line), so you can see what's going on. To answer your last question: you can apply the above deformation retract to map one of those Möbius bands to the circle. In particular the orange line, which is your $A\cap B$, gets mapped to the circle by this deformation retract.