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

Question

Let $M,N$ be smooth manifolds and $F:N\mapsto M$ a diffeomorphism. Prove that for every $p\in N$

$(dF^{-1})_{F(p)}=(dF)^{-1}_p$

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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 ?

Answer 1

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 $$d(F^{-1})\big|_{F(p)} = (dF\big|_p)^{-1} \equiv (dF_p)^{-1}.$$

Answer 2

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 $F$.