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What is difference between annulus (cylinder) and disk in graph routing?

I know annulus is disk with hole, or I can imagine how is similar to cylinder, but my problem is, I can't understand this paragraph:

A fundamental difference between the disc and cylinder problems is that in the case of the cylinder there is more than one homotopically inequivalent route between pairs of points of $bd(\Sigma)$, even if we restrict ourselves to non-self-intersecting routes, which we can.

I can't understand why we have more than one different route between two points on annulus but one route on disk? (I think in both cases we should have one route).

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up vote 1 down vote accepted

If you have a simple path between two points in the disk, you can morph it continuously into any other path in the disk, so homotopically all paths in the disk are equivalent.

However, in an annulus, there are paths that cannot be morphed continuously into each other (while keeping the same endpoints - the hole gets in the way), so they aren't equivalent. Think about paths that wind around the hole vs paths that don't.

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So homotopically equivalent means I should be able to convert one to another by continues moves, yes? could you provide sample which makes life harder by this property of annulus? (actually any problem which is easier on disk but harder on annulus, and is easy to see this) – Saeed Jan 14 '13 at 18:05
@Saeed You can define a $\sqrt{}$ function on any disk in $\mathbb C$ that avoids $0$. But you cannot define a $\sqrt{}$ function on every annulus that avoids $0$. – Hagen von Eitzen Jan 14 '13 at 18:11

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