Fixed point property of Mobius strip

Question

Does the mobius strip has fixed point property? If it isn't what is the map from mobius strip to itself witch lacks fixed point?

Answer 1

The Möbius band $M$ does not have the fixed point property.

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For an intuitive understanding, take a look at this wikipedia movie. Imagine, instead of a crab on $M$, just a single point $p \in M$. Move $p$ a small distance $s$ in the circular direction, taking one little crab step. Doing that simultaneously for each $p$ defines a fixed point free homeomorphism of $M$.

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Here's a more rigorous construction.

First, $M$ can be expressed as the quotient of the topological space $\mathbb R \times [-1,+1]$ under the equivalence relation $$(x,t) \sim \bigl(x+n, \,(-1)^n t\bigr) \quad\text{for all $n \in \mathbb Z$} $$ This is just a variation of the more familiar quotient space description of $M$, as the quotient of $[0,1] \times [-1,+1]$ under the equivalence relation $(0,t) \sim (1,-t)$;

Consider, for each $s \in \mathbb R$, the homeomorphism $$f_s : \mathbb R \times [-1,+1] \to \mathbb R \times [-1,+1] $$ defined by the formula $$f_s(x,t) = (x+s,t) $$

Now verify for yourself that this homeomorphism preserves equivalence classes in the sense that $$(x,t) \sim (x',t') \iff f_s(x,t) \sim f_s(x',t') $$ It follows, by the universal property of quotient maps, that $f$ induces a homeomorphism $$F : M \to M $$ Furthermore if $s \not\in \mathbb Z$ then $(x,t) \not\sim (x+s,t)$ for all $(x,t) \in \mathbb R \times [-1,+1]$, and therefore $F$ has no fixed points.