How does null-homotopic loops exist in torus?

Question

If we identify torus by $S^{1} \times S^{1}$, then what are the ways null-homotopic loops exist in torus? I don't see a way to do this except we walk around the genrator back and forth. Since if we consider two genrators a,b for the torus, then any loop is generated by a,b, but these loops will all go around the torus, which means they will detect holes, so those loops are not null-homotopic? I seem to have a misunderstanding for this part...Any help would be appreciated.

Answer 1

If you remove the red circle below, you will get an open cylinder, e.g. something diffeomorphic to $(0,1)\times S^1$. Then the purple circle will have image in the resulting cylinder corresponding to the line segment $(0,1)\times \{\theta\}$ for some $\theta \in S^1$. Removing that line segment will leave you with a rolled up copy of $\Bbb{R}^2$, as indeed $S^1$ minus a point is diffeomorphic to $(0,1)$, and $(0,1)\times (0,1)\cong \Bbb{R}^2$.

The upshot of this is that as long as your loop does not intersect the union of the red and purple circles below, it will be null homotopic.

Edit: image used from wikipedia: http://en.wikipedia.org/wiki/Torus#Geometry

Answer 2

Take a walk along the generators back and forth and obtain a trivial loop. An easier way is take a point and a small ball around it, then the boundary of the ball is null homotopic.

Answer 3

You probably know about the universal covering map $$f : \mathbb R \times \mathbb R \to S^1 \times S^1 $$ which is a continuous map defined by $$f(x,y) = (e^{2\pi i x}, e^{2 \pi i y}) $$ (using complex notation for the unit circle $S^1 \subset \mathbb C$).

You probably also know that every loop in $\mathbb R \times \mathbb R$ is null homotopic.

So, consider any loop whatsoever in $\mathbb R \times \mathbb R$, namely ANY continuous function $\gamma : [0,1] \to \mathbb R \times \mathbb R$ with $\gamma(0)=\gamma(1)$.

Since $\gamma$ is null-homotopic in $\mathbb R \times R$, and since $f$ is continuous, it follows that $f \circ \gamma$ is null homotopic in $S^1$ times $S^1$

It follows that the loop $f \circ \gamma : [0,1] \to S^1 \times S^1$ is null homotopic in $S^1 \times S^1$, because of $H : [0,1] \times [0,1] \to \mathbb R \times R$ is a null-homotopy of $\gamma$ than $f \circ H$ is a null-homotopy of $f \circ \gamma$.

Notice, $\gamma$ could be a terrible function, as long as its continuous and $\gamma(0)=\gamma(1)$. It could even be one of those terrible Peano "space filling" curves whose image is an entire solid square $[0,1] \times [0,1]$, which case the curve $f \circ \gamma : [0,1] \to S^1 \times S^1$ would be surjective. And yet, $f \circ \gamma$ is null-homotopic.

Finally, if you understand the theory of universal covering maps, in particular the general lifting lemma from covering space theory, then you will know that this construction I gave is actually a construction of all possible null homotopic closed curves in $S^1 \times S^1$.