Prove that for every $p\in N$ we have $(dF^{-1})_{F(p)}=(dF)^{-1}_p$, with $F:N\to M$ a diffeomorphism Let $M,N$ be smooth manifolds and $F:N\mapsto M$ a diffeomorphism. Prove that for every $p\in N$
$(dF^{-1})_{F(p)}=(dF)^{-1}_p$

Foremost because $F$ is  a diffeomorphism $\text{dim}N=\text{dim}M$ and the matrix $(dF)_p$ is invertible.
My idea was to take the matrix  $(dF^{-1})_{F(p)}$ and do the multiplication with $(dF)_p$, if the final product is $I_n$ then the inverse is $(dF^{-1})_{F(p)}$ as we wanted.
Here is where I stuck:
$\left [ \frac{\partial (F^i)^{-1}}{\partial x^j} \right ]\cdot \left [ \frac{\partial (F^i)}{\partial x^j} \right ]=$
\begin{bmatrix}
\frac{\partial (F^1)^{-1}}{\partial x^1}\cdot \frac{\partial (F^1)}{\partial x^1}+\frac{\partial (F^1)^{-1}}{\partial x^2}\cdot \frac{\partial (F^2)}{\partial x^1}+...+\frac{\partial (F^1)^{-1}}{\partial x^n}\cdot \frac{\partial (F^n)}{\partial x^1} &.  &.  &.  &\frac{\partial (F^1)^{-1}}{\partial x^1}\cdot \frac{\partial (F^1)}{\partial x^n}+...+\frac{\partial (F^1)^{-1}}{\partial x^n}\cdot \frac{\partial (F^n)}{\partial x^n} \\ 
 .&  .&  &  & \\ 
 .&  &  .&  & \\ 
 .&  &  &  .& \\ 
 \frac{\partial (F^n)^{-1}}{\partial x^1}\cdot \frac{\partial (F^1)}{\partial x^1}+...+\frac{\partial (F^n)^{-1}}{\partial x^n}\cdot \frac{\partial (F^n)}{\partial x^1}&  .&  .&  .& \frac{\partial (F^n)^{-1}}{\partial x^1}\cdot \frac{\partial (F^1)}{\partial x^n}+...+\frac{\partial (F^n)^{-1}}{\partial x^n}\cdot \frac{\partial (F^n)}{\partial x^n}
\end{bmatrix}
in order for the above to be $I_n$ it  must be true that $\frac{\partial (F^i)^{-1}}{\partial x^i}\cdot \frac{\partial (F^i)}{\partial x^i}=\frac{\partial (F^i)^{-1}\circ F^i }{\partial x^i}=1$ and the rest should be zeros
but I dont think this is correct, I have never seen $\frac{\partial (F^i)^{-1}}{\partial x^i}\cdot \frac{\partial (F^i)}{\partial x^i}=\frac{\partial (F^i)^{-1}\circ F^i }{\partial x^i}=1$
Can someone clarify this for me, if my idea is correct how should I proceed ? and if it's not can you give a hint ?
 A: One the one hand, we have $d(F^{-1})_{F(p)}$, which is the differential of $F^{-1}:M\to N$, calculated at $F(p)\in M$. This is a map of the form $d(F^{-1})_{F(p)}:T_{F(p)}M\to T_p N$, and more precisely the one acting on tangent spaces $[\gamma]\in T_{F(p)} M$ as (I'm using parentheses to stress the order of the operation and who's who's input):
$$\color{red}{(} d(F^{-1})_{F(p)} \color{red}{)}([\gamma])
= [F^{-1}\circ \gamma] \in T_p N,$$
On the other hand, consider $dF_p^{-1}\equiv (dF_p)^{-1}$. More precisely, this is the inverse of $dF_p:T_p N\to T_{F(p)}M$, and thus $(dF_p)^{-1}:T_{F(p)}M\to T_p N$.
Using the properties of the differential operation, we have $$d(F^{-1})\circ dF = d(F^{-1}\circ F)=d(\text{id}_N) = \text{id}_{TN},
\\
dF\circ d(F^{-1}) = \text{id}_{TM.}$$
It follows that $d(F^{-1}):TM\to TN$ is the inverse operation of $dF:TN\to TM$. Considering their restricted action to specific fibers, this then in particular means that $d(F^{-1})\big|_{F(p)}\equiv (dF^{-1})_{F(p)}$ is the inverse of $dF\big|_p\equiv dF_p$. Which can be equivalently expressed writing that
$$d(F^{-1})\big|_{F(p)} = (dF\big|_p)^{-1} \equiv (dF_p)^{-1}.$$
A: This follows from the functoriality of $dF_p$. I will denote $dF_p$ by $F_{*,p}$ or $F_{*}$. Once you prove the following data:
$1.$ If $id:M\rightarrow M$ denotes the identity then for each $p\in M$, $id_{*,p}=id_{T_pM}$
$2.$ $(F\circ G)_{*}=F_{*}\circ G_{*}$
You obtain the result you seek by taking $G$ to be the inverse of $F$.
