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I'm trying to create a function from $\mathbb R^2 \rightarrow \mathbb R^2$ such that the image of quadrant 1 $(+x, +y)$ is precisely quadrants 1 and 3 or the pre-image of quadrants 1 and 3 is precisely quadrant 1.

I don't really have any idea how to go about deriving such a function, I've currently tried a variety of maps that look like $f(x,y)=\frac {x}{x-y},\frac{y}{x-y}$ and other things that look a bit like that. So far, nothing quite works out the way I am hoping for it too.

They don't have to actually be a bijection so long as the number of places that the function doesn't map to is of measure zero.

Does anyone have any ideas for how a function like this might look, or at least ideas to try? I feel like maybe some mappings that are a bit weirder might work here, but I'm not sure what to even try.

Alternatively, if someone could give an explanation for why this might actually be impossible that would be great! I've got no luck in that direction either.

amWhy
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Vibbz
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1 Answers1

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In polar coordinates, the first quadrant can be written as $$Q_1 = \{ (r\cos(\theta), r\sin(\theta)) : r>0, 0 \leq \theta < \pi/2\}.$$ We can write $Q_1 = Q_1^+ \cup Q_1^-$, where $$Q_1^- = \{ (r\cos(\theta), r\sin(\theta)) : r>0, 0 \leq \theta < \pi/4\},$$ $$Q_1^+ = \{ (r\cos(\theta), r\sin(\theta)) : r>0, \pi/4 \leq \theta < \pi/2\}.$$ These sets are disjoint. Visually, we have bisected the first quadrant. Now, we are going to define a function which maps the lower bisected region to all of the first quadrant and maps the upper bisected region to all of the third quadrant. Define $f:Q_1 \to \mathbb{R}$ to be $$ f(x,y) = f (r\cos(\theta), r\sin(\theta)) = (r\cos(2(\theta)), r\sin(2(\theta))), \quad x,y\in Q_1^-, $$ and $$ f(x,y) = f (r\cos(\theta), r\sin(\theta)) = (-r\cos(2(\theta - \pi/4)), -r\sin(2(\theta - \pi/4))), \quad x,y\in Q_1^+. $$ The function $f$ is defined on all of $Q_1^-\cup Q_1^+ = Q_1$, is injective, and is onto. Thus it is invertible.

Matt E.
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  • It needs to be defined from all of R^2, though. This function is only defined from the first quadrant. How do you extend it to the other 3 quadrants? – Vibbz Mar 25 '22 at 16:19