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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?

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  • $\begingroup$ 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). $\endgroup$ – Severin Schraven Oct 13 at 20:42
  • $\begingroup$ Yes, I see that this is enough. But how can I link $(f_*)_p$ with the given differential? $\endgroup$ – user661541 Oct 13 at 20:52
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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.

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