A "bunch" of time?? Really, you Americans crack me up sometimes!

Anyway, I'm not sure I know the full answer to your question. Let's see.....

By our definition of a manifold (locally homeomorphic to some

), we see that

is

*trivially* a manifold. You may take this to mean that "locally" is redundant, and the definition is "trivially" satisfied.

Now

is a manifold, locally homeomorphic to the 2-cylinder, which is just a "rolled up" 2-plane (homeomorphic to it, iow). But it is also globally homeomorphic to it, so, by the above, the definition is trivially satisfied, albeit at one remove, as it were.

As to why the local homeomorphism is called "local triviality" I really cannot say - I just think of it as being a definition. Maybe that's wrong of me, dunno. Looking back,I see I didn't explain it very well - in fact I erred, which I will fix. Let's try again.

Recall I said there is a projection

and that the preimage

, the fibre over

. Suppose

is a neighbourhood in

. Then I should find that

is the

**bundle** over

ie a local bundle. By the assertion that

*locally* , I intended

, and not what I first said.

*blush*

Anyway let's look at our recipe for a general fibre bundle

. This a manifold of dimension

, and will be called a fibre bundle if the following hold:

there is a base manifold

of dimension

a typical fibre

of dimension

(all fibres being identical)

a projection

such that for

, the fibre over

.

a structure group

characteristic to

local triviality given by

.

I need to say something more about our structure group, but first we will need to think about orientation. Now you will find several definitions out there, but we are just going to go with intuition. Let's stick with our 1-sphere.

Let's assume we know what meant by (anti-)clockwise for a vanilla circle, and suppose that for any arc-segment of the circle, this still makes some sort of sense, Then, when we think of the 1-sphere as a manifold, these arc-segments are coordinate neighbourhoods. We will say that two such neighbourhoods are

*consistently oriented* if they are both oriented (anti-)clockwise. If

*all* neighbourhoods are consistently oriented, we will say are manifold is

**orientable**.

Now since our tangent vectors at, say

are simply directional derivatives of the coordinate there, it is easy to see the tangent space "inherits" its orientation form the neighbourhood.

Armed with that we are ready to proceed, but I need to pause for breath a while