# Inverse function theorem and tangent space differentials

Having some difficulty with the following proposition,

Consider a $$C^\infty$$ map between manifolds $$F:N \rightarrow M$$ with $$\text{dim} \, N = \text{dim} \, M$$, then $$F$$ is locally invertible at some point $$p \in N$$ if $$F_{*,p} : T_pN \rightarrow T_{F(p)}M$$ is an isomorphism.

Now, I think I understand the general idea of how to show this, but am stuck on a finer point. Of course, the pushforward of a basis vector on $$T_pN$$ given charts about $$p \in N$$ and $$F(p) \in M$$, $$(U, x^1,...,x^n)$$ and $$(V,y^1,...,y^n)$$ respectively, $$F_{*,p} \bigg( \frac{\partial}{\partial x^i} \bigg|_p\bigg) = \sum_j^n \frac{\partial F^j}{\partial x^i}(p) \frac{\partial}{\partial y^j} \bigg|_{F(p)} \tag{1}$$ with $$F^j := y^j \circ F$$. My intuition is to recall that for a linear operator $$T: \mathbb{R}^n \rightarrow \mathbb{R}^n$$ such that $$T(v) = A v$$ for $$v \in \mathbb{R}^n$$ and $$A \in \text{GL}(n,\mathbb{R})$$, then $$T$$ is invertible iff $$T$$ is an isomorphism, in the hopes that similar logic can be applied to more abstract vector space transformations.

So, naturally, I want to argue that the pushforward of $$\partial / \partial x^i |_p$$ is represented by the matrix $$\partial F^j / \partial x^i |_{F(p)}$$, and the differential being an isomorphism is equivalent to the inverse matrix existing (which means that $$\text{det}[\partial F^j / \partial x^i |_{F(p)}] \neq 0$$), and thus $$F$$ is invertible via the inverse function theorem on manifolds.

But $$F_{*,p}$$ is then acting in a form like $$T(v) = A w = z$$, with $$v \in T_pN$$ and $$w,z \in T_{F(p)}M$$. I'm unable to prove an analogue of the variant with $$T$$ on $$\mathbb{R}^n$$, because I can't just naively toss around $$A^{-1}$$ on $$z$$ to get an element in $$T_pN$$, as $$w$$ is in $$T_{F(p)}M$$! The linear combination is of basis vectors $$\partial / \partial y^j$$, and not $$\partial / \partial x^i$$, which exist in different spaces, no? How can I prove that $$F_{*,p}$$ is an isomorphism $$\iff \text{det}[\partial F^j / \partial x^i |_{F(p)}] \neq 0$$ to complete the proof?

Remember that $$\frac{\partial F^j}{\partial x^k}(p):=\frac{\partial G^j}{\partial r^k}\phi (p)$$ for $$G:=\psi\circ F\circ \,\phi^{-1}$$ where $$\phi:=(x^1,\ldots ,x^n),\,\psi :=(y^1,\ldots ,y^n)$$ and $$r:=(r^1,\ldots ,r^n)$$ is the Cartesian coordinate system in $$\mathbb{R}^n$$. Then as $$\phi$$ and $$\psi$$ are local diffeomorphisms of $$M$$ and $$N$$ respectively with $$\mathbb{R}^n$$ then $$G:\mathbb{R}^n\to \mathbb{R}^n$$ is an smooth map and $$\det\left[\frac{\partial F^j}{\partial x^k}(p)\right]=\det\left[\frac{\partial G^j}{\partial r^k}\phi (p)\right]$$ Therefore if the determinant is not zero the derivative of $$G$$ at the point $$\phi(p)$$, that is $$G_{*,\phi(p)}$$, is an automorphism of $$\mathbb{R}^n$$ and so it is invertible. Now observe that $$\phi_*$$ and $$\psi_*$$ are isomorphisms (because they are assumed to be diffeomorphisms) and by the chain rule we have that $$F_{*,p}=\psi _{*,F(p)}^{-1}\circ G_{*,\phi(p)}\circ \phi _{*,p}$$, then as $$F_{*,p}$$ is composition of invertible linear maps its invertible.
By the other hand, if the determinant is zero then $$G_{*,\phi(p)}$$ is not invertible and so it kernel is not trivial, so there is some $$x\in \mathbb{R}^n\setminus \{0\}$$ such that $$G_{*,\phi(p)}x=0$$, and so $$F_{*,p}((\phi_{*,p}) ^{-1}x)=0$$, so $$F_{*,p}$$ is not invertible because, being $$\phi_*$$ an isomorphism, $$(\phi_{*,p}) ^{-1}x\neq 0$$ when $$x\neq 0$$.∎
• Oh, I see, I forgot to refer to the definition of what it means for $F$ to be $C^{\infty}$ in terms of charts. Thanks much! Commented Jan 6, 2021 at 23:58