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
[Math] Why are the Möbius strip and the boundary of a Klein bottle homotopy equivalent to $S^1$
algebraic-topologyklein-bottlemobius-band
Related Solutions
This crude drawing may be helpful.
We can consider the Klein Bottle $K$ to be the quotient space of the cylinder $C = S^1 \times I$ by the identification $(z,0) \sim (z^{-1},1)$. Here we consider $S^1$ as the unit circle in the complex plane. Let $L = 1 \times I$. I claim that $K$ retracts onto the loop $L'$ defined by $L$ under the quotient map.
Let $q : C \to K$ be the quotient map. The map $q$ satisfies the following universal property:
If $g : C \to Z$ is a continuous map from $C$ to any topological space $Z$ such that $a \sim b$ implies $g(a) = g(b)$ for all $a,b \in C$ then there exists a unique map $f : K \to Z$ such that $g = f \circ q$.
Define a map $g : C \to S^1$ as follows: First project onto $L$, then quotient to $L'$. This is a continuous map which is preserved under $\sim$, hence a continuous map $f : K \to L'$ exists and must be the identity on $L'$ by the universal property.
Best Answer
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.