Let M be the Mobius band. Prove there is no retract from M to its boundary.
Can I use fundamental group stuff here?
The below lemma can be found in the standard topology/algebraic topology books.
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Lemma 1. If A is a retract of X, then the homomorphism of fundamental groups induced by inclusion $\displaystyle j:A \rightarrow X$ is injective.
Proof: If $\displaystyle r:X \rightarrow A$ is a retraction, then the composite map $\displaystyle r \cdot j$ equals the identity map of A. It follows that $\displaystyle r_* \cdot j_* $ is the identity map of $\displaystyle \pi_1(A,a)$, so that $\displaystyle j_*$ must be injective, where $\displaystyle r_*$ and $\displaystyle j_*$ denote the induced homomorphism by topological continuous maps r and j.
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A deformation retract of a Mobius band is its middle circle. A degree 1 map of a circle involves a degree 1 map of a middle circle of a Mobius band. Thus, the induced map $\displaystyle j_*:\pi(S^1) \rightarrow \pi(M)$ in Lemma 1 is an isomorphism.
However, a Mobius band is a non-orientable two-dimensional manifold (surface). A two dimensional figure in a Mobius band cannot be moved around the middle circle once and back to where it started. Actually, coming back to the initial point of a Mobius band involves two turns of its middle (or inner) circle. If the boundary were to be a retract of a Mobius band, it is homotopic to the degree 2 map of a middle circle to itself.
A degree 1 map of a circle to a subset A of a Mobius band and degree 2 map of a circle of a subset B of a Mobius band cannot induce the same homomorphisms (isomorphisms) between their fundamental groups. Since the middle circle is a deformation retract of a Mobius band, the boundary of a Mobius band cannot be the deformation retract of a Mobius band. It follows that $\displaystyle r_* \cdot j_* $ in lemma 1 is not the identity map if the boundary were to be a retract of a Mobius band.
Thus, there is no retract from a Mobius band to its boundary.