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Let $M$ a submanifold of $\mathbb R^n$, for all $x$ in $M$, let $\pi_x:\mathbb R^n\rightarrow T_xM$ the orthogonal projection onto the tangent space $T_xM$ of $M$ at $x$.
How could you show that for all $x$ in $M$, it exists an open neighborhood $U$ of $x$, such as for all $y$ in $U\cap M$, $\pi_y$ restricted on $U$ is a diffeomorphism from $U$ onto $\pi_y(U)$?

Thank you.

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Use \mathbb{R}^n for $\mathbb{R}^n$. – M.B. Jun 2 '12 at 21:36
How can a diffeomorphism exist between $U$ (dimension $n$) and $T_yM$ (dimension strictly smaller than $n$)? Also, $\pi_y$ is a linear map, so the statement you're trying to prove sounds a bit weird IMHO. – Giuseppe Negro Jun 2 '12 at 22:33
I assume that the neighborhood is being taken with respect to $M$. – Dylan Moreland Jun 2 '12 at 22:42
It seems very plausible indeed. For small enough $U$ you can just ''flatten'' $U$ to fit onto its image in the tangent space. And $U$ is small enough such that the map is injective. – M.B. Jun 2 '12 at 23:01
The derivative of $\pi_y$ has full rank everywhere on $U$ as long as the normal space at $y$ has zero intersection with the tangent space at $x$. This is sure to happen if the angle between tangent spaces at $x$ and $y$ is small. So I guess one can begin by arguing (or recalling) that the map $y\mapsto T_yM$ (from $M$ to the Grassmannian) is continuous... – user31373 Jun 2 '12 at 23:55
up vote 2 down vote accepted

Change the coordinates such as $x=0$ and $\mathbb R^n=T_xM\oplus V=\{(t,w)\}$, so $T_xM:w=0$.
By the local inverse function theorem there is an open neighborhood of $x$ such as $\pi_x:U\rightarrow\pi_x(U)$ is a diffeomorphism.
Denote by $f:V=\pi_x(U)\rightarrow U$ the inverse function. So $f$ is a local parameterization of $M$ at $x$ by a neighborhood of $0$ in $T_xM$. We have $f(0)=0$ and $d_xf=0$ (since $T_xM:w=0$).

Let $y=(t,f(t))\in U\cap M$, then $T_yM=\{(t+h,f(t)+d_tf(h))\}$.
It exists $\varepsilon>0$ such as $\|d_tf\|<\varepsilon$ for all $(t,f(t))\in U$, so $p_y$ is injective on $U$.
Let $z\in U$, we check that $d_z(p_y):T_zU\rightarrow T_yU=T_yM$ coincides with the orthogonal projection $T_zU\rightarrow T_yM$ : $p_y:U\hookrightarrow R^N\rightarrow T_yM$ so $d_z(p_y)=T_zU\hookrightarrow R^N\rightarrow T_yM$.So $d_z(p_y)$ is invertible.
So $p_y$ is a diffeomorphism on $U$.

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