Tangent Vectors and Differentials
Definition 1. A tangent vector at $p\in M$ is defined as a linear functional $v: C ^{\infty}(M) \rightarrow \mathbb{R}$, which satisfies: $$ v(fg) = f(p)vg+g(p)vf, \forall f, g\in C ^{\infty}(M)$$ The linear space $T_p M$ of all tangent vectors at $p$ is defined as the tangent space at $p$.
When $M = \mathbb{R}^n$, we can see that tangent vectors are just derivative operators along some directions: $$\frac{\partial}{\partial e}\bigg{|}_p fg = f(p) \frac{\partial}{\partial e}g(p) + g(p)\frac{\partial}{\partial e} f(p)$$ where $$ \frac{\partial}{\partial e}\bigg{|}_p f := \frac{d}{d t}\bigg{|} _{t=0}f(p+te) = e^i \frac{\partial }{\partial x^i} f(p)$$ And $\{\frac{\partial }{\partial x^i}\big{|}_p\}$ is the basis of $T_p \mathbb{R}^n$.
Definition 2. The differential of $F: M\rightarrow N$ at $p\in M$ is a linear map: $dF_p: T_p M \rightarrow T _{F(p)}N$ defined as: $$dF_p(v)(f) = v(f\circ F), \forall v\in T_p M, \forall f\in C^{\infty}(N)$$
The definitions above tell us that we need to use smooth functions to “feel” the tangent spaces and differentials.
Same as smoothness, tangent vectors are local.
Proposition 1. Let $v\in T_p M$. If $f, g\in C^{\infty}(M)$ agree on some neiborhood $U$ of $p$, then $vf = vg$
Proof. $h = f-g$ is zero at $p$. Find a bump function $\psi$ for $\text{supp }h$ suppported in $M\setminus\{p\}$, so $\psi h \equiv h$. Then $v(h) = v(\psi h) = 0$
Proposition 2. Let $U\subset M$ be open and $\imath: U \hookrightarrow M$ is the inclusion map. Then for each $p \in M$, $d\imath_p :T_p U \rightarrow T_p M$ is an isomorphism.
Proof. $d\imath_p$ is a linear map, so we only need to prove that it is bijective. For injectivity, we choose a neighborhood $B$ of $p$ such that $\overline{B}\subset U$. Assume $d \imath_p(v) = 0$. Then for any $f\in C^{\infty}(U)$, we can find $\tilde{f}\in C^{\infty}(M) $ that agrees with $f$ on $\overline{B}$. So $v(f) = v(\tilde{f}\circ \imath) = d \imath_p(v) (\tilde{f}) = 0$. For surjectivity, for any $w\in C^{\infty}(M)$, we define $v\in C^{\infty}(U)$ by $vf = w\tilde{f}$. Then we need to check $d\imath_p(v) = w$: $d\imath_p(v)f = v(f\circ i) = w(\widetilde{f\circ i}) = wf$ (because $\widetilde{f\circ i}$ and $f$ agree on $B$).
For boundary points, we define tangent space by the following lemma:
Lemma 1. Let $\imath: \mathbb{H}^n\hookrightarrow \mathbb{R}^n$, for each $a\in \partial \mathbb{H}^n$, $d\imath_a: T _a \mathbb{H}^n\rightarrow T_a\mathbb{R}^n$ is an isomorphism.
The scheme of proof is as the same as proposition 2: $f\in C^{\infty}(\mathbb{H}^n)$ can be extended to defined on all $\mathbb{R}^n$.
Proposition 3. The dimension of $T_p M$ at any $p$ is $n$.
Proof. Let $(U, \phi)$ be a chart containing $p$, then $d\phi_p$ is an isomorphism between $T_p U$ and $T _{ \hat{p} } \hat{U}$. $dim(T _{ \hat{p} } \hat{U}) = n$.
Coordinate Representation
To compute with tangent vectors or differentials, we need to use coordinates.
Definition 3. Le $(U, \phi)$ be a local chart containing $p$, we define $$ \frac{\partial}{\partial x^i}\bigg{|}_p f = \frac{\partial}{\partial x^i}\bigg{|} _{\phi(p)} \hat{f} = \frac{\partial\hat{f}}{\partial x^i}(\hat{p}).$$ It can be checked that $\frac{\partial}{\partial x^i}\big{|}_p$ is a tangent vector at $p$.
In the representation of coordinates, we can represent the basis of $T_p M$.
Proposition 4. $\{ \frac{\partial} {\partial x^i} \big{|}_p\} _{i=1}^n$ is a basis of $T_pM$.
Every element of $T_p M$ has an representation of the form $$v = v^i \frac{\partial }{\partial x^i}$$
For a differential of a smooth map:
$$ \begin{split}&dF_p(\frac{\partial}{\partial x^i}\bigg{|}_p)(f) = \frac{\partial}{\partial x^i}\bigg{|}_p(f\circ F) \cr = &\frac{\partial f}{\partial y^j}(F(p))\frac{\partial F^j}{\partial x^i}(p) = \left(\frac{\partial F^j}{\partial x^i}(p) \frac{\partial}{\partial y^j}\bigg{|} _{F(p)}\right)f\end{split}$$
So the matrix representation of $dF_p$ is $\left(\frac{\partial F^i}{\partial x^j}(p)\right) _{n\times m}$, which is called the Jacobian matrix of $F$. The rank of this matrix is defined as the rank of $F$.
Maps of Constant Rank
Defintion 4. If the rank of $F: M \rightarrow N$ is a constant $r$ on all points of $U$, we say $F$ has constant rank on $U$. If $r = dim(N)$, we call $F$ a smooth submersion. If $r = dim(M)$, we call $F$ a smooth immersion, and if $F$ is a topological embedding, we call it a smooth embedding.
Since the determinant function is continuous, if at some $p\in M$, $rank(F) \not = 0$, then we can find a neighborhood $U$ of $p$, such that $F$ has a constant rank on $U$.
By the inverse function theorem on $\mathbb{R}^n$ and coordinate representation of smooth maps, we can prove similar theorem for manifolds:
Theorem 1 (Inverse Function Theorem for Manifolds). Suppose $M$ and $N$ are smooth manifolds without boundary, $F:M\rightarrow N$ is a smooth map. If $dF_p$ is invertable at $p$ , then there exist connected neighborhoods $U_0$ centered at $p$ and $V_0$ of $F(p)$ such that $F\big{|} _{U_0}$ is a diffeomorphism from $U_0$ to $V_0$.
This theorem may fails for manifolds with boundaries. Consider $\imath: \mathbb{H}^n \hookrightarrow \mathbb{R}^n$, $d\imath_p$ is isomorphism on $\partial \mathbb{H}^n$. But $\mathbb{H}^n$ is not diffeomorphic to $\mathbb{R}^n$ locally on the boundary.
Defintion 5 (Local Diffeomorphism). Let $F: M\rightarrow N$ be a smooth map. If $\forall p \in M$, there exists a neighborhood $U$ such that $F\big{|}_U$ is a diffeomorphism.
According to inverse function theorem, if $M, N$ are manifolds without boundary, $F: M\rightarrow N$ is a local diffeomorphism if and only if $dF_p$ is invertable at every $p\in M$.
Rank Theorem
Theorem 2 (Rank Theorem). Suppose $M$ is a $m$-manifold and $N$ is a $n$-manifold, without bounary, $F:M\rightarrow N$ is a smooth map with constant rank $r$, then $\forall p\in M$, there exist smooth chart $(U, \phi)$ centered at $p$ and $(V, \psi)$ centered at $F(p)$, such that $F(U) \subset V$ and in which $F$ has a coordinate representation as follows: $$\hat{F}(x^1, …, x^m) = (x^1, …, x^r, 0, …, 0)$$
Proof. This theorem is local, so we can use smooth charts to replace $M, N$ by $U\in \mathbb{R}^m, V \in \mathbb{R}^n$. Now we permutate the order of components of $\mathbb{R}^m$ such that $DF_p$ has a non-singular $r \times r$ upper left submatrix. We write coordinates in $\mathbb{R}^n$ as $(x, y) = (x^1, …, x^r, y^{1}, …, y^{m-r})$ and coordinates in $\mathbb{R}^n$ as $(v, w) = (v^1, …, v^r, v^{1}, …, v^{n-r})$. We assume $p = (0,0)$ and $F(p) = (0,0)$. We write $$F(x, y) = (v, w) = (Q(x,y), R(x, y)),$$ then $(\frac{\partial Q^i}{\partial x^j})$ is non-singular at $(0,0)$. We define $$\phi(x, y) = (Q(x, y), y) \in \mathbb{R}^m.$$ We can check that $D\phi(0,0)$ is invertable. So by the inverse function theorem, there exist connected neighborhoods $U_0, \tilde{U_0}$ of $(0,0)$ such that $\phi \big{|} _{U_0}$ is a diffeomorphism from $U_0$ to $\tilde{U_0}$. We assume $$\phi^{-1}(x, y) = (A(x,y), B(x, y)),$$ then $$(x, y) = \phi \circ \phi^{-1}(x, y) = \phi(A(x,y), B(x, y)).$$ We get $B(x,y) = y$ and $Q(A(x, y), y) = x$. So $$\begin{split}F\circ \phi ^{-1}(x, y) = &F(A(x, y), y) = (Q(A(x, y),y), R(A(x, y), y))\cr = & (x, \tilde{R}(x,y))&\end{split}.$$ We shrink $U_0$ such that $F$ has aconstant rank $r$, so $\frac{\partial \tilde{R}^i}{\partial y^j} = 0$, that is, $\tilde{R}(x, y) = S(x)$ Now we only need to find a coordinate transformation $\psi$ of $\mathbb{R}^n$ to eliminate $S(x)$. To do this, we define $\psi(v, w) = (v, w-S(v))$. We can check that $D\psi(0,0)$ is non-singular. And we have $$\psi \circ F \circ \phi^{-1}(x, y) = (x, 0)\ \ on\ U_0.$$
If $M$ has a boundary and $N$ does not has, we have the following rank theorem:
Theorem 3 (Local Immersion Thoerem for Boundaries). Suppose in addition $M$ has a boundary and $p\in \partial M$, then there exist a boundray chart $(U, \phi)$ centered at $p$ and a interior chart $(V, \psi)$ centered at $F(p)$ such that $F(U) \subset V$ and $\hat{F}(x) = (x, 0)$.
Proof. We extend $F$ near $p$ as before. By rank theorem, we can find $(U_0, \phi_0)$ and $(V_0, \psi_0)$ such that $\psi_0 \circ F \circ \phi_0 ^{-1}(x) = (x, 0)$. But $(U_0, \phi_0)$ may not restricted to be a boundary chart. Now we define $\psi = (\phi_0^{-1}, Id _{\mathbb{R}^{n-m}})\circ \psi_0$. Then $$\psi\circ F (x)= \psi\circ F \circ \phi^{-1} \circ \phi(x) = (x, 0).$$
Reference
Introduction to Smooth Manifolds