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If we have a function $F(x): \mathbb{R^4} \rightarrow \mathbb{R^3}$. Defined as \begin{align} x_1\, x_4&=y_1 \\ x_2\, x_4&=y_2 \\ x_1^2+x_2^2-x_3^2&=y_3 \end{align} Can a continuous inverse map exist? I'm intuitively guessing that the problem with continuity can occur at $y=(0,0,1)$ but I'm not being able to prove it.

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  • $\begingroup$ You might have to go for parts of $\mathbb{R}^4$ instead of the full domain. $\endgroup$ – mvw Aug 6 '15 at 15:13
  • $\begingroup$ @mvw I endorse your comment as a way to make this question most interesting. $\endgroup$ – James S. Cook Aug 6 '15 at 16:01
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A continuous inverse map cannot exist because an inverse map cannot exist, as the function is not injective. For example $$F(1,0,0,0)=(0,0,1)=F(0,1,0,0).$$

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  • $\begingroup$ f(1,0,0,0)=f(-1,0,0,0)=(0,0,1). So we do not know how to go back. $\endgroup$ – Adelafif Aug 6 '15 at 15:04
  • $\begingroup$ Thanks a lot for your comment! Could we possibly obtain a continuous function that could give us a value of x for every value of y such that F(x)=y even if a unique inverse was not available due to injectivity? $\endgroup$ – Leo Euler Aug 6 '15 at 20:02
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The question as asked is easier than what I attack here. Certainly the lack of injectivity spells doom for an inverse, much less a continuous inverse. That said, it's fun to think about how we could suitably restrict the given function as to obtain an inverse. The Jacobian matrix essentially reveals when and what we can do in that regard. I give an illustration of this below:

Observe, $F(x_1,x_2,x_3,x_4) = (y_1,y_2,y_3)$ defined by: \begin{align} x_1\, x_4&=y_1 \\ x_2\, x_4&=y_2 \\ x_1^2+x_2^2-x_3^2&=y_3 \end{align} for all $(x_1,x_2,x_3,x_4) \in \mathbb{R}^4$ has Jacobian matrix: $$ J_F = \left[ \frac{\partial F}{\partial x_1} \bigg{|} \frac{\partial F}{\partial x_2}\bigg{|}\frac{\partial F}{\partial x_3}\bigg{|}\frac{\partial F}{\partial x_4}\right] = \left[ \begin{array}{cccc} x_4 & 0 & 0 & x_1 \\ 0 & x_4 & 0 & x_2 \\ 2x_1 & 2x_2 & -2x_3 & 0 \end{array}\right]$$ This shows us what the dimension of the image is locally. In particular, the rank of the Jacobian shows us the dimension of the image near the point as the component functions are polynomial and hence continuous. For example, if $x_3=0$ then we need the following determinant to be nonzero in order that $J_F$ have rank 3: $$\text{det}\left[ \begin{array}{ccc} x_4 & 0 & x_1 \\ 0 & x_4 & x_2 \\ 2x_1 & 2x_2 & 0 \end{array}\right] = -2x_4(x_2^2+x_1^2)$$ which is nonzero for $x_4 \neq 0$ and $(x_1,x_2) \neq (0,0)$. With this in mind, I return to the original system of equations and invert them with respect the given conditions: \begin{align} x_1\, x_4&=y_1 \\ x_2\, x_4&=y_2 \\ x_1^2+x_2^2 &=y_3 \end{align} we wish to solve for $x_1,x_2,x_4$ in terms of $y_1,y_2,y_2$. Note, $x_1/x_2 = y_1/y_2$ thus $ y_1^2+y_2^2 = x_4^2y_3$. Therefore, \begin{align} x_4 &= \pm \sqrt{(y_1^2+y_2^2)/y_3} \\ x_1 &= y_1/x_4 = \pm y_1/\sqrt{(y_1^2+y_2^2)/y_3} \\ x_2 &= y_2/x_4 = \pm y_2/\sqrt{(y_1^2+y_2^2)/y_3} \end{align} The formulas above give a pair of local inverses for $F$ restricted to the three-dimensional subset $S$ of $\mathbb{R}^4$ for which $(x_1,x_2,x_3,x_4)$ has $x_3=0$ and $x_4 \neq 0$ and $(x_1,x_2) \neq (0,0)$. In particular, if we denote $S=S_+ \cup S_-$ where $S_+$ has $x_4 >0$ and $S_-$ has points with $x_4<0$ then the formulas above define inverse functions for $F$ restricted to $S_\pm$.

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  • $\begingroup$ incidentally, it is worthwhile to remember, any function can be modified into a bijection by choosing a cross-section of the domain which hits each fiber of the function just once and by setting the range to be the codomain. Moreover, this is generally far from a unique process! $\endgroup$ – James S. Cook Aug 6 '15 at 16:00
  • $\begingroup$ Thanks a lot for your comment! What I had in mind while asking was could we possibly obtain a continuous function that could give us a value of $x$ for every value of $y$ such that $F(x)=y$ even if a unique inverse was not available due to injectivity? $\endgroup$ – Leo Euler Aug 6 '15 at 17:52
  • $\begingroup$ Glad to help, incidentally, what I claim here is a natural outgrowth of the inverse or implicit function theorem proof. You might get some ideas from en.wikipedia.org/wiki/… also, it's still a work in progress, but Chapter 5 of supermath.info/AdvancedCalculus13.pdf may be useful to you. $\endgroup$ – James S. Cook Aug 6 '15 at 18:42
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Continuous bijections between $ \mathbb R^m \rightarrow \mathbb R^n$ ( bijection is needed for the existence of a global inverse ) are not possible unless $n=m$: otherwise, restrict the domain to a closed, bounded ( so compact )ball) B .Then $ h_|B \rightarrow h(B) $ is a continuous bijection between compact and Hausdorff, a homeomorphism, which cannot happen between subsets of $\mathbb R^n , \mathbb R^m ; n \neq m$ by, e.g., invariance of domain.

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