# Inverse function theorem for manifolds

Let $$f: M \to N$$ be a smooth map and $$(f_*)_p: T_pM \to T_{f(p)}N$$ an isomorphism. Then there exists an open neighborhood $$W$$ of $$p$$ such that $$f\vert_W: W \to f(W)$$ is a diffeomorphism.

Here $$(f_*)_p$$ is the map that sends a tangent vector $$[\gamma] \mapsto [f \circ \gamma]$$. My notes says that this immediately follows from the inverse function theorem that I'm used to, but I don't see how it follows.

How can I prove this?

Attempt:

Take charts $$(U_p, \phi_p)$$ around $$p$$ and $$(V_{f(p)}, \psi_{f(p)})$$ around $$f(p)$$ in the smooth atlases of the manifolds with $$f(U_p) \subseteq V_p$$. We then know that $$\psi_{f(p)} \circ f \circ \phi_p^{-1}$$ is $$C^\infty$$ at $$\phi(p)$$. How can I proceed?

• All you need to check is that the differential $D (\psi_{f(p)} \circ f \circ \phi_p^{-1})$ is an isomorphism at the point $\phi(p)$. Then you know that $\psi_{f(p)} \circ f \circ \phi_p^{-1}$ is a diffeo around $\phi(p)$ by the inverse function theorem. This then implies your claim (as your charts are diffeos). – Severin Schraven Oct 13 at 20:42
• Yes, I see that this is enough. But how can I link $(f_*)_p$ with the given differential? – user661541 Oct 13 at 20:52

Assume $$\text{dim}\ M=\text{dim}\ N=n.$$ In what follows, we use the Einstein convention for all sums.

The point is that if $$(U_p, \phi_p)$$ is a chart about $$p$$ in $$M$$ and $$(V_{f(p)}, \psi_{f(p)})$$ is a chart about $$f(p)$$ in $$N$$ then $$(f_*)_p: T_pM \to T_{f(p)}N$$ is a linear transformation, so it has a matrix representation in the coordinates defined by $$\phi$$ and $$\psi$$. If we can show that this matrix is the Jacobian of $$\hat f:=\psi_{f(p)} \circ f \circ \phi_p^{-1}$$, which is $$\left(\frac{\partial \hat f^j}{\partial r^i}\right)_{ij},$$ then $$\hat f$$ will be a local diffeomorphism, which is what we want.

But, $$\textit{by definition},\ \frac{\partial }{\partial x^i}=(\phi_*)^{-1}\frac{\partial }{\partial r^i}$$, where $$(r^i)$$ are the usual Euclidean coordinates. Similarly, $$\frac{\partial }{\partial y^i}=(\psi_*)^{-1}\frac{\partial }{\partial s^i}$$ where we use $$(s^i)$$ to represent the Euclidean coordinates in the range of $$\hat f$$ just to make the calculations easier to follow. For the same reason, we drop the subscripts $$p$$ and $$f(p).$$ Finally, we note that by the chain rule in $$\mathbb R^n,\ \hat f_*\frac{\partial }{\partial r^i}=\frac{\partial \hat f^j}{\partial r^i}\frac{\partial}{\partial s^j},$$ where the $$(\hat f^j)$$ are the components of $$\hat f$$. Then, we calculate

$$f_*\frac{\partial }{\partial x^i}=f_*\circ (\phi_*)^{-1}\frac{\partial }{\partial r^i}=(f\circ \phi^{-1})_*\frac{\partial }{\partial r^i}=$$

$$(\psi^{-1}\circ \hat f)_*\frac{\partial }{\partial r^i}=(\psi_*)^{-1}\circ \hat f_*\frac{\partial }{\partial r^i}=$$

$$(\psi^{-1})_*\frac{\partial \hat f^j}{\partial r^i}\frac{\partial}{\partial s^j}=\frac{\partial \hat f^j}{\partial r^i}(\psi^{-1})_*\frac{\partial}{\partial s^j}=\frac{\partial \hat f^j}{\partial r^i}\frac{\partial}{\partial y^j}$$.

It follows that the matrix of $$f_*$$ is the Jacobian of $$\hat f,$$ as desired.