Prove that there is not a retraction of the Möbius band to its boundary

Question

I would like a hint to solve the following problem:

Let $X$ be the Möbius band and $A$ its boundary (which is a circumference). Prove that there is not a retraction $r:X \rightarrow A$.

Each try I did gave no solution.

Any hint to solve this problem?

Thanks!

Answer 1

The fundamental group of a Moebius band is just $\mathbb{Z}$ (you can see that as it is homotopy equivalent to a circle). Pick a loop that generates it. If you don't know much about the fundamental group, just pick a circle that runs through the middle of the Moebius band; notice, it should go around once, not twice like the boundary.

Now, if there was a retraction this loop should map to a loop in the boundary. Is that possible? Why not? Can the class of the loop change in a retraction?

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Edit: Extra hint:

What happens if you pre-compose with the inclusion $A \rightarrow X$ ? There are two maps induced $$\mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}$$ What do we know about the first, and what do we know about their composition?