[matheist]

Sorry to actually your actually, but the derivative of a function f from a space A to a space B at the point a is a linear function Df_a from the tangent space of A at a to the tangent space of B at b = f(a).

When the spaces are Euclidean spaces then we conflate the tangent space with the space itself because they're identical.

By the way, this makes it easy to remember the chain rule formula in 1 dimension. There's only one logical thing it could be between spaces of arbitrary dimensions m, n, p: composition of linear transformations from T_a A to T_f(a) B to T_g(f(a)) C. Now let m = n = p = 1, and composition of linear transformations just becomes multiplication.

(Only half kidding)

[btilly]

The distinction between the space A and the tangent space of A becomes visually clear if we consider a function whose domain is a sphere. The derivative is properly defined on the tangent plane, which only touches the sphere at a single point. However in the neighborhood of that point, the plane and sphere are very, very close together. But are inevitably pulled away by the curvature of the sphere.

Of course that picture is not formally correct. We formally define the tangent space without having to embed the manifold in Euclidean space. But that picture is a correct description of an embedding of both the sphere and the tangent space at a single point.

[ziofill]

Oh I appreciate you actualling my actually ^^ but isn’t this case a special case of the one I wrote? I.e. when an and b are manifolds and admit tangent bundles?

[beng-nl]

Why, I’m sure you could come up with a succinct explanation of a monad :-)