Tangent space for a surface. He shows why the tangent space ($T_p S$) is a vector space at every point. Composed of Tangent vectors

{Tangent space for {a surface defined as a level set of a function}} is equal to {the kernel of the Differential at each point of the surface} (note that the differential is a Linear map, and the kernel is the subspace which the map maps to 0). vid

General definition on manifolds require defining the vector space of derivatives! Formal definition of tangent vector and space – From the calculation we can see that it corresponds to what we would intuitively think, the vectors can be just be constructed in the usual way by just thinking in the domain of the chart. and Tangent space can be written as span of these vectors, it could depend on the choice of Coordinate chart.. but we can prove it doesn't. Proof that both definitions agree.

In the general definition it is defined as a differential operator, defined to be a Linear map $t_x: C^\infty (M) \to \mathbb{R}$ which satisfies the Leibinz rule:

$t_x(fg) = f(x) t_x(g) + t_x(f) g(x)$

It turns out that this forms a Vector space, whose basis vectors can be $\frac{\partial}{\partial x_i}$ for a particular set of coordinates

It is the Dual vector space to the Cotangent space

Tangent vectors transform as Contravariant Tensors. This can be seen because for a smooth map between manifolds, the Differential gives the transformation between tangent spaces:

$DF_a (\frac{\partial}{\partial x_i})_a (f) = \sum_j \frac{\partial F_j}{\partial x_i} (a) (\frac{\partial}{\partial y_j})_{F(a)} (f)$

A reparametrization is a particular case of such a map.

The tangent space is used to define the Differential.

There is a Functor associating to each smooth map between differentiable manifolds, a map between the corresponding tangent spaces. This functor is Covariantly functorial, so that it gives a map between the tangent space at doamin to that at codomain. See discusion in Cotangent space