How to proof that closed path $l$ (see picture) on 2-dimensional torus don't homotopic to trivial path, using definition torus as CW-complex? (closed path on surface $S$ is a map $f:[0, 1]\to S$, such that $f(0)=f(1)=x_0$) Thanks.
Closed path on torus
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0It is difficult to give an answer that will be useful to _you_, without knowing your mathematical background. It would also be helpful to know where the question comes from. Is this for a class? Self-study? Do you have someone to talk to? What approaches have you tried? – 2011-09-15
1 Answers
Four proof sketches. $\newcommand{\ZZ}{\mathbb Z} \newcommand{\RR}{\mathbb R}$
Note that the vertical loop (that you drew) and the horizontal loop have a single intersection point. Since the parity of the number of transverse intersection points is invariant under homotopy neither the vertical nor horizontal are homotopically trivial.
The fundamental group of the torus is $\ZZ \times \ZZ$. The vertical and horizontal loops give the generators and so neither is trivial.
Since the torus is the quotient $\RR^2/\ZZ^2$ we can take the preimage of the vertical loop in $\RR^2$ and get a collection of vertical lines. By homotopy lifting, a trivial loop in the torus lifts to an infinite collection of trivial loops in $\RR^2$.
A loop in a surface is homotopically trivial if and only if it bounds a disk. The vertical loop is non-separating, so in particular does not bound a disk.