I'm going to say some of the same things, but in a different way.
A manifold is merely something that is
locally homeomorphic to Euclidean space. (Some people allow the different connected components of a manifold to be locally homeomorphic to Euclidean spaces of different dimension)
From there, we would then like to say that a differentiable manifold is merely something that is
locally diffeomorphic to Euclidean space. However, this raises a problem: what the heck is a differentiable map??
But let's ignore that for the moment.
In Euclidean space, we like to talk about things like functions, and vector fields.
That's easy enough for a manifold, but we can do something even more interesting!
When considering functions, it is often useful to think of their
graphs. For example, a function
R->
C is simply a subset of
Rx
C for which there is exactly one point on each vertical line.
Let's call these vertical lines
fibers. (Or
fibres) We then say that
Rx
C is the
trivial fiber bundle over the
base space R with
fiber C.
A
section of a fiber bundle simply assigns to each point in the base space a single element of the corresponding fiber. So, the notion of a function
R->
C coincides with the notion of a section of the fiber bundle
Rx
C. (We usually like to talk about continuous sections, which corresponds to a continuous function)
Now, let's consider real-valued functions of the circle S^1.
It's easy to picture the trivial bundle S^1 x
R: it's simply a cylinder (or annulus, if you prefer). But it may strike you that there's a more interesting way to form a fiber bundle: you could "twist" the fibers as you go around the circle.
In other words, the Möbius strip is a fiber bundle over S^1, with fiber isomorphic to
R!
The salient feature is that any point of S^1 has a neighborhood U in which the corresponding collection of fibers together look like the trivial fiber bundle UxF.
The difference between these two fiber bundles is interesting! one way to tell them is that any continuous section of the Möbius strip must have a zero, whereas the same is not true of the cylinder.
It turns out, I think, that these are the only two fiber bundles over S^1 with fiber
R. (Another interesting thing to study!)
If we recall in our work with Euclidean space, one of the important aspects of differentiability is that of tangent vectors. Or more generally, things like vector fields that assign a vector to each point in space. So, this suggests we might like to talk about a
vector bundle (a fiber bundle whose fibers are vector spaces) which we'll call the
tangent bundle.
T
R^n is simply the trivial bundle
R^n x
R^n. (The first component is the base point, and the second component is the tangent vector)
Then, we can relate all of the tangent bundles for Euclidean spaces. For example, if we have a differentiable map f:
R² ->
R³, we have the induced map on the tangent bundles f_*:T\mathbb{R}^2 \rightarrow T\mathbb{R}^3 given by:
f_*(x, v) := (f(x), (Df)(x) v)
(Df)(x) v is, of course, simply the directional derivative along v.
We define a
differentiable manifold to be a vector bundle that is compatable with all of this structure we've placed on the Euclidean spaces.
(This is the "abstract nonsense" approach -- most introductions will actually construct such beasts, and then later prove that it's compatable with this structure)
So now that we have a differentiable structure, we probably want to do calculus! Thus, differential geometry!