Fixed point property of Mobius strip 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?
 A: The Möbius band $M$ does not have the fixed point property.

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 $M$.

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.
