Explication of the differential of a smooth map

Solution 1:

The differential (or tangential map) is a map between tangent bundles induced from a smooth function between manifolds, such that its linear in every tangent space at a point of the manifold. It can be represented locally using the dual basis of the local basis in use of the tangent space, that is, suppose you have a chart $(U,\varphi )$ for a smooth manifold $M$ of dimension $n$, this means that $U\subset M$ is open and $\varphi :U\to \mathbb{R}^n$ is a diffeomorphism.

Now, $\varphi $ defines a frame on $TU$, this means that if $\varphi =(\varphi _1,\ldots ,\varphi _n)$ for smooth real-valued functions $\varphi _k$ then we can use the symbols $\frac{\partial}{\partial \varphi ^1}\big|_p,\ldots ,\frac{\partial}{\partial \varphi ^n}\big|_p$ to represent a basis of the tangent space of every point $p\in U$. These symbols represent the same than the canonical basis of $T_p\mathbb{R}^n$, but they represent vectors in $T_pU$ instead of in $T_p\mathbb{R}^n$.

Using these symbols you can do standard multivariable calculus as if you will be in $\mathbb{R}^n \cong T_p\mathbb{R}^n$, but this just work locally (this is the point of the whole differential geometry). Now, using the frame of above you can represent the differential $d\varphi :TU\to T\mathbb{R}^n$ using the dual basis of $\frac{\partial}{\partial \varphi ^1},\ldots ,\frac{\partial}{\partial \varphi ^n}$, that is, you define a basis $d\varphi ^1,\ldots ,d\varphi ^n$ on the dual space of $TU$ such that $d\varphi ^k\left(\frac{\partial}{\partial \varphi ^j}\right)= \delta _{k,j}$, where the latter is the Kronecker $\delta $.

Now, any smooth function $f: M\to N$ defines a differential that can be represented in $U$ using the basis $d\varphi ^1,\ldots ,d\varphi ^n$ induced by the chart $\varphi $, that is, you will have that $df=\sum_{k=1}^n c_k d\varphi ^k$ for some smooth functions $c_k:M\to N$.

If $df=0$ it means that the differential map any vector on $TM$ to the corresponding zero vector in $TN$. As tangent vectors in $TM$ represent derivations you are mapping any derivation in $M$ to the zero derivation in $N$.