Let $U$ be an open subset of $\Bbb{R}^{2}$ and

\begin{align*}\psi:U&\rightarrow\Bbb{R}^{4}\\ x\mapsto &(\psi_1(x),\psi_2(x),\psi_3(x),\psi_4(x)) \end{align*} a $\mathcal C^1$ function.

Assume that there exists $(\overline{x_1},\overline{x_2})\in U$ such that $\text{d}\psi_{(\overline{x_1},\overline{x_2})}$ is injective.

First, I needed to prove the existence of $i,j\in\{1,2,3,4\},$ $i<j$ and an open set $V\subset U$ containing $(\overline{x_1},\overline{x_2})$, such that, the function $h:V\rightarrow\Bbb{R}^{2}$ defined by


is a diffeomorphism of class $\mathcal C^1$ into an open set $W\subset \Bbb{R}^{2}$.

I have proved the above statement using the inverse function theorem. Note that, if $\mathrm{d}\psi_{(\overline{x_1},\overline{x_2})}$ is injective, we can choose two rows $ i,j$ of this matrix that are linearly independent, so we can apply the inverse function theorem to $h$.

But the second question is where I'm stucked:

Consider $i=1,j=2$. I need to show that $\forall$ $(x_1,x_2)\in V,$ $ \exists!$ $(y_1,y_2)\in W$ and a function $f=(f_1,f_2):W\rightarrow \Bbb{R}^{2}$ such that $$\psi(x_1,x_2)=(y_1,y_2,f(y_1,y_2)).$$

My problem here is how to prove that $\dfrac{\partial \psi}{\partial x_3,x_4}$ is non-singular.

By the same idea of the first question, the linearly independents rows of $\mathrm{d}\psi_{(\overline{x_1},\overline{x_2})}$ are $1$ and $2$.

What can I do?


1 Answer 1


What you can do is back off from the problem, and try to look at it again sometime in the future with fresh eyes. You've made some errors in interpreting the problem and are currently on a wild goose chase.

There is no need for $\dfrac{\partial \psi}{\partial x_3,x_4}$ to be non-singular. There are no restrictions on $\psi$'s behavior with respect to $x_3, x_4$ at all. Using your earlier notation $(y_1, y_2) = h(x_1, x_2)$ and $$f(y_1, y_2) = (\psi_3\circ h^{-1}(y_1, y_2),\psi_4\circ h^{-1}(y_1, y_2))$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .