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This is a problem from Arnold's book on ODEs I cannot solve.

Prove that if $f:U\to V$ is a diffeomorphism, then the Euclidean spaces with the domains $U$ and $V$ as subsets have the same dimension. Hint. Use the implicit function theorem.

Thanks.

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    Hint: If an n x m matrix is invertible, what can you say about the matrix?2010-12-26

1 Answers 1

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Thanks for the comments. After giving it more thought I was able to solve it on my own. Here's what I did.

Suppose $U\subset \mathbb{R}^{n+m},V\subset \mathbb{R}^m$ and $n>0$ (if $n<0$ just consider $f^{-1}$ instead of $f$ in what follows). We have

$f'=\left( \begin{array}{cccccc} \frac{\partial f_1}{\partial x_1} & \ldots & \frac{\partial f_1}{\partial x_n} & \frac{\partial f_1}{\partial y_1} & \ldots & \frac{\partial f_1}{\partial y_m} \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ \frac{\partial f_m}{\partial x_1} & \ldots & \frac{\partial f_m}{\partial x_n} & \frac{\partial f_m}{\partial y_1} & \ldots & \frac{\partial f_m}{\partial y_m} \end{array} \right)=\left( \begin{array}{cc} \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(x_1,\ldots ,x_n\right)} & \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)} \end{array} \right)$

$\left(f^{-1}\right)'=\left( \begin{array}{ccc} \frac{\partial x_1}{\partial f_1} & \ldots & \frac{\partial x_1}{\partial f_m} \\ \ldots & \ldots & \ldots \\ \frac{\partial x_n}{\partial f_1} & \ldots & \frac{\partial x_n}{\partial f_m} \\ \frac{\partial y_1}{\partial f_1} & \ldots & \frac{\partial y_1}{\partial f_m} \\ \ldots & \ldots & \ldots \\ \frac{\partial y_m}{\partial f_1} & \ldots & \frac{\partial y_m}{\partial f_m} \end{array} \right)=\left( \begin{array}{c} \frac{\partial \left(x_1,\ldots ,x_n\right)}{\partial \left(f_1,\ldots ,f_m\right)} \\ \frac{\partial \left(y_1,\ldots ,y_m\right)}{\partial \left(f_1,\ldots ,f_m\right)} \end{array} \right)$

$\left(f^{-1}\right)'\cdot f'=\left( \begin{array}{cc} \frac{\partial \left(x_1,\ldots ,x_n\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(x_1,\ldots ,x_n\right)} & \frac{\partial \left(x_1,\ldots ,x_n\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)} \\ \frac{\partial \left(y_1,\ldots ,y_m\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(x_1,\ldots ,x_n\right)} & \frac{\partial \left(y_1,\ldots ,y_m\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)} \end{array} \right)=\left( \begin{array}{cc} I_{(n\times n)} & 0 \\ 0 & I_{(m\times m)} \end{array} \right)=I_{(n+m\times n+m)}$

It follows that

$\frac{\partial \left(y_1,\ldots ,y_m\right)}{\partial \left(f_1,\ldots ,f_m\right)}\cdot \frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)}=I_{(m\times m)}$

Since the Jacobian matrix $\frac{\partial \left(f_1,\ldots ,f_m\right)}{\partial \left(y_1,\ldots ,y_m\right)}$ is not singular, we can apply the implicit function theorem. It follows that there is a function $g:A\to B$, where $A\subset \mathbb{R}^n,B\subset \mathbb{R}^m$ are open subsets and $A\times B\subset U$, such that $f(x,g(x))=\text{const}$ for all $x\in A$. It is then clear that $f$ cannot be bijective.

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    I'm very happy to see that you posted this an an answer to your own question! See [this old post](http://meta.math.stackexchange.com/questions/1277/a-suggestion-for-dealing-with-users-who-post-homework-problems).2010-12-26