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Supposed that the derivative of $f:X\to Y$ is an isomorphism whenever $x$ lies in the sub-manifold $Z \subset X$, and assume that $f$ maps $Z$ diffeomorphically onto a $f(Z)$ . Prove that $f$ map a neighborhood of $Z$ diffeomorphically on to a neighborhood of $f(Z)$

Here is what I got so far

Since the derivative of $f:X\to Y$ is an isomorphism whenever $x$ lies in the sub-manifold $Z \subset X$, for $x\in Z$, $f$ is locally diffeomorphism. and since $f$ maps $Z$ diffeomorphically onto a $f(Z)$, there is a local inverse $g_i: U_i \to X$ where $U_i$ is a locally finite collection of opensubset of $Y$ covering $f(Z)$

Let $W=\{y\in U_i:g_i(y)=g_j(y)$ whenever $y\in U_i \cap U_j\}$ Since we have $f$ is locally diffeomorphism, $g_i$ can be patch together to define a smooth inverse $g:U\to X$. According to the partition of unity properties, $g_i$ is finitely many. I know that I need to show that $W$ contain an open neighborhood of $f(Z)$, but my brain cant find any way to do so. Maybe in my previous arguments, I missed something

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Let $Z$ be a compact submanifold of the Euclidean space $\mathbb{R}^d$. Let $f:T^{\perp}Z\to\mathbb{R}^d$ be \begin{eqnarray} f\left(c^i\left(\dfrac{\partial}{\partial x^i}\right)_x\right)=x+\left( \begin{array}{ccc} c^1\\ \vdots\\ c^d \end{array}\right). \end{eqnarray} $Z'$ denotes the zero section of $T^{\perp}Z$.

Then, $df_x$ is isomorphc whenever $x\in Z'$ and $f$ maps $Z'$ diffeomorphically onto $Z$.

In the figure below, the point $y\in Z$ satisfies that $y\in U_1\cap U_2$. But \begin{eqnarray} g_2(y)=0\in T_y Z\\ g_1(y)=\xi^i\left(\dfrac{\partial}{\partial x^i}\right)_{y'}, where\\ y-y'=\left(\begin{array}{ccc} \xi^1\\ \vdots\\ \xi^d \end{array}\right) \end{eqnarray}. So the point $y$ doesn't belong to $W$. Therefore even $Z\subset W$ doesn't hold in your construction of $W$.

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  • $\begingroup$ so how do you think I should modify $W$ to make this work? $\endgroup$ – Diane Vanderwaif Oct 13 '14 at 17:00

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