# Path homotopy in two-holed torus

I'm reading Lee's Introduction to topological manifolds and on page 291 he writes about the two-holed torus $X$:

"In terms of our standard generators for $\pi_1(X)$ this loop is path-homotopic to either $\alpha_1 \beta_1 \alpha_1^{-1} \beta_1^{-1}$ or $\beta_2 \alpha_2 \beta_2^{-1} \alpha_2^{-1}$ so it is not null homotopic..."

where he is talking of the path around the middle bit as for example on this picture http://inperc.com/wiki/images/b/ba/Double_torus_construction.jpg the red path in the third part of the picture.

I'm confused about this because according to my understanding this path is null homotopic because I can move it to one side of the two-holed torus and then shrink it to a point as there are no holes inside it.

Can someone explain to me why this is false? Many thanks for your help!

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I understand that it's not easy to explain this in writing, but what do you mean by "move it to one side of the two-holed torus"? Because if you want simply move the loop to the left, the left hole will block you at some point. – Martin Sleziak Jul 25 '11 at 12:28
Imagine miniature people on the two-hole torus standing side-by-side, all holding hands, forming the given red circle. They can all march however they want, but they must always hold hands and gravity keeps their feet glued to the surface. Do you expect them to just supernaturally hover across over the donut holes? Magical antigravity powers would sure make homotopy trivial. :) – anon Jul 25 '11 at 12:39
@anon: Ha :-O! Best explanation ever! Now it seems blatantly obvious, thank you! – Rudy the Reindeer Jul 25 '11 at 12:46