Rank theorem on manifolds says that :
Suppose $M$ and $N$ are two smooth manifolds of dimensions $m$ and $n$, respectively, and $F:M\rightarrow N$ be a smooth map with constant rank $r$. For each $p\in M$ there exists a smooth chart $(U,\varphi)$ around $p$ and a smooth chart $(V,\psi)$ around $F(p)$ such that $F(U)\subseteq V$ and $$\psi\circ F\circ \varphi^{-1}:\varphi^{-1} (U)\subseteq \mathbb{R}^m\rightarrow\psi(V)\subseteq \mathbb{R}^n$$ is given by $(a_1,\cdots,a_m)=(a_1,\cdots,a_r,0,\cdots,0)$.
There are many books where proof of this has been discussed but none of the books I have seen has proof that I feel excited. So, I am trying to produce another proof which I think is most natural.
Let $p\in M$. As $F$ is a map of constant rank, $dF_p:T_pM\rightarrow T_{F(p)}(N)$ is of rank $r$. A Linear algebra results says that, in this case, there exists basis for $T_pM$ and a basis for $T_{F(p)}(N)$ such that $dF_p:T_pM\rightarrow T_{F(p)}N$ is represented by matrix $\begin{bmatrix}I_r&0\\0&0\end{bmatrix}$ i.e., it is given by $$(v_1,\cdots,v_r,v_{r+1},\cdots,v_m)\rightarrow (v_1,\cdots,v_r,0,\cdots,0).$$ I am almost sure that the basis of $T_pM$ and the basis of $T_{F(p)}N$ that we choose above corresponds to charts around $p$ and $F(p)$ respectively. I will try to elaborate what I said. How do we think of a basis of tangent space at a point?
Given $p\in M$, we take any chart $(U,\varphi)$ around that point given by $\varphi=(x_1,\cdots,x_n)$ and then see that $$\left\{\frac{\partial}{\partial x_i}\bigg|_p:1\leq i\leq n\right\}$$ is a basis for $T_pM$. Now I am hoping to trace back. Given a basis of $T_pM$ can we trace back to obtain a chart $(U,\varphi)$ at $p$. Suppose we could find that charts $(U,\varphi)$ at $p$ and $(V,\psi)$ at $F(p)$ such that the corresponding basis at tangent spaces is the choice of basis that we have made above.
Let $(U,\varphi=(x_1,\cdots,x_m))$ be chart based at $p$. Then $\{\partial/\partial x_i|_p:1\leq i\leq m\}$ is the basis of $T_pM$ that we have mentioned above and similarly $(V,\psi=(y_1,\cdots,y_n))$ be chart based at $F(p)$ and $\{\partial/\partial y_j|_{F(p)}:1\leq i\leq n\}$ is the basis of $T_{F(p)}N$ that we have mentioned above. So, we have $$dF_p(a_1,\cdots,a_m)=(a_1,\cdots,a_r,0,\cdots,0).$$ So, $$dF_p\left(\sum_{i=1}^ma_i\frac{\partial}{\partial x_i}\bigg|_p\right) =\sum_{i=1}^ra_i\frac{\partial}{\partial y_i}\bigg|_{F(p)}$$ $$\sum_{i=1}^ma_idF_p\left(\frac{\partial}{\partial x_i}\bigg|_p\right) =\sum_{i=1}^ra_i\frac{\partial}{\partial y_i}\bigg|_{F(p)}$$ $$\sum_{i=1}^ma_i\left(\sum_{k=1}^n\frac{\partial F^k}{\partial x_i}\frac{\partial}{\partial y^k} \bigg|_{F(p)}\right) =\sum_{i=1}^ra_i\frac{\partial}{\partial y_i}\bigg|_{F(p)}$$ Here I have assumed $F$ to be a map (locally) from $\mathbb{R}^m\rightarrow \mathbb{R}^n$ and $F^k$ are the corresponding functions of $F$.
I still do not see how to use this and conclude that $\psi\circ F\circ \varphi^{-1}(a_1,\cdots,a_m)=(a_1,\cdots,a_r,0,\cdots,0).$
Am I going in correct way? Can you suggest something to complete this proof