Commutativity of a diagram involving differential of a smooth map and a Jacobian matrix

Suppose that $$M$$ and $$N$$ are smooth manifolds of dimension $$m$$ and $$n$$ respectively, and that $$F: M \to N$$ is a smooth map. Fix $$p \in M$$, and suppose that $$(U, \varphi)$$ and $$(V, \psi)$$ are coordinate charts containing $$p$$ and $$F(p)$$, respectively.

Let $$T_p M$$ be the set of all derivations of $$C^\infty(M)$$ at $$p$$, $$T_{F(p)} N$$ to be the set of all derivations of $$C^\infty(N)$$ at $$F(p)$$, and let $$\widehat{F}$$ be the coordinate representation of $$F$$ with respect to the charts $$(U, \varphi)$$ and $$(V, \psi)$$. Then $$\varphi$$ induces an ordered basis $$\left(\frac{\partial}{\partial x^1}\bigg|_p, \frac{\partial}{\partial x^2}\bigg|_p, \dots, \frac{\partial}{\partial x^n}\bigg|_p\right)$$ on $$T_p M$$ where \begin{align} \frac{\partial}{\partial x^i}\bigg|_p : C^\infty(M) \to \mathbb{R}, f \mapsto \frac{\partial}{\partial x^i}\bigg|_p(f \circ \varphi^{-1}), \end{align} for all $$1 \leq i \leq n$$.

Similarly, $$\psi$$ induces an ordered basis $$\left(\frac{\partial}{\partial x^1}\bigg|_{F(p)}, \frac{\partial}{\partial x^2}\bigg|_{F(p)}, \dots, \frac{\partial}{\partial x^n}\bigg|_{F(p)}\right)$$ on $$T_{F(p)} N$$, where the $$\frac{\partial}{\partial x^j}\bigg|_{F(p)}$$ are defined in a similar manner to the $$\frac{\partial}{\partial x^i}\bigg|_p$$, using $$\psi$$. Let $$\alpha: T_p M \to \mathbb{R}^n$$ be the coordinate map with respect to the ordered basis $$\left(\frac{\partial}{\partial x^i}\bigg|_p\right)_{1 \leq i \leq n}$$ on $$T_p M$$, and $$\beta: T_{F(p)} N \to \mathbb{R}^m$$ be the coordinate map with respect to the ordered basis $$\left(\frac{\partial}{\partial x^i}\bigg|_{F(p)}\right)_{1 \leq i \leq m}$$ on $$T_{F(p)} N$$.

My question is, does the following diagram commute? If so, why?

$$\require{AMScd}$$ $$\begin{CD} T_p M @>{dF_p}>> T_{F(p)} N\\ @V{\alpha}VV @VV{\beta}V\\ \mathbb{R}^n @>{J_{\varphi(p)} \widehat{F}}>> \mathbb{R}^m \end{CD}$$ Here $$J_{\varphi(p)} \widehat{F}$$ denotes the linear map from $$\mathbb{R}^n$$ to $$\mathbb{R}^m$$ given by the $$m \times n$$ Jacobian matrix of $$\widehat{F}$$ at $$\varphi(p)$$.

I am trying to state this question in general terms, but I am mainly interested in the case where $$N = M$$ and $$M$$ has a global coordinate chart $$(M, \varphi)$$.

Another reason why I have asked this question is that I would like to see how the concept of the differential of a smooth map relates to the concept of the Jacobian matrix of a smooth map between Euclidean spaces.

• Observe that $$\beta \circ dF_p \circ \alpha^{-1} (v^1,\dots,v^m)^T = \Big(\sum_i v^i \partial_i \hat{F}^1(\hat{p}), \dots, \sum_i v^i \partial_i \hat{F}^n(\hat{p}) \Big)^T = J_{\hat{p}} \hat{F} (v^1,\dots,v^m)^T.$$ Aug 18 '20 at 13:58

It does: in the case $$M=\mathbb{R}^m$$ and $$N=\mathbb{R}^n$$, if I write $$dF_p:T_p\mathbb{R}^m\to T_{F(p)}\mathbb{R}^n$$ the differential of $$F$$ as a map between derivations and $$DF_p:\mathbb{R}^m\to\mathbb{R}^n$$ the usual derivative of $$F$$, then the map $$dF_p$$ sends $$\partial_v|_p$$ on $$\partial _{DF_p(v)}|_{F(p)}$$, where $$\partial_w|_q$$ is the directional derivative in the direction $$v$$ at the point $$p$$, namely $$\partial_w|_qf=\lim\limits_{h\to0}\frac{f(q+hw)-f(q)}{h},$$ which is what the commutativity of your diagram means.
To see it, we use the fact (coming from usual calculus) that for a function $$g:\mathbb{R}^p\to\mathbb{R}$$, we have $$\partial_v|_pg=Dg_p(v)\,\,(\ast),$$
and the chain rule: thus, for a smooth function $$f:\mathbb{R}^n\to\mathbb{R}$$, we get
$$dF_p(\partial_v|_p)f:=\partial_v|_p(f\circ F)\overset{(\ast)}{=}D(f\circ F)_p(v)\overset{c.r.}{=}Df_{F(p)}(DF_p(v))\overset{(\ast)}{=}\partial_{DF_p(v)}|_{F(p)}f.$$
Since they are equal on all such functions $$f$$, the derivations $$dF_p(\partial_v|_p)$$ and $$\partial_{DF_p(v)}|_{F(p)}$$ are equal. Thus, $$dF_p$$ sends $$\partial_v|_p$$ on $$\partial_{DF_p(v)}|_{F(p)}$$.