Show $f:\mathbb{R}^2 \rightarrow \mathbb{R}^{2}, f(x,y) = (x-y, x^2-y^2)$ [closed]

Show $f:\mathbb{R}^2 \rightarrow \mathbb{R}^{2}, f(x,y) = (x-y, x^2-y^2)$ is a bijection

This is my try $f$ is injective if $$ (a,b) \neq (x,y) \rightarrow f(a,b) \neq f(x,y)$$

If $$f(x,y)=f(a,b) = a \qquad \forall a,b,x,y\in \mathbb{R}^{2}$$

Therefore is not injective

For the surjective

$f$ is surjective if $$Im(f)= \mathbb{R}^2 \quad \forall a, b \in \mathbb{R}^{2} \in (x,y) \in \mathbb{R}^{2}: (a,b) = f(x,y)$$

Please someone guide me to the correct solution.

Solution 1:

This is false. The function is neither surjective nor injective.

Not surjective. Consider $(0, 1) \in \mathbb{R}^2$. There's no $(x,y) \in \mathbb{R}^2$ such that $f(x,y) = (0, 1)$. Why? In the first coordinate, we have $x-y = 0$. But then, the second coordinate is $$ x^2-y^2 = (x-y)(x+y) = 0\,(x+y) = 0 \neq 1. $$

Not injective. By evaluation, $f(0,0) = (0,0) = f(1,1)$, but $(0,0) \neq (1,1)$.

Edited to show how to prove a closely related function is a bijection, inspired by (now deleted) answer by Brian Chao.

This formula does define a bijective function $g: A \to B$, where $A$ and $B$ are each a subset of $\mathbb{R}^2$ formed by removing a particular line: \begin{align} A &= \{ (x,y) \in \mathbb{R}^2 \mid x \neq y \}, \\ B &= \{ (u,v) \in \mathbb{R}^2 \mid u \neq 0 \} \end{align} We are using the same formula, but the domain and range are different, so this is a different function.

This function $g$ has a (two-sided) inverse $h: B \to A$, defined by $$ h(u, v) = \biggl( \frac12 \Bigl(\frac{v}{u} + u \Bigr),\, \frac12 \Bigl(\frac{v}{u} - u \Bigr) \biggr) $$ It's a good exercise to verify that they are mutually inverse, i.e. $h\bigl(g(x,y)\bigr) = (x,y)$ as long as $x \neq y$ and $g\bigl(h(u,v)\bigr) = (u,v)$ as long as $u \neq 0$.

Why does the existence of an inverse show that $g$ is a bijection (and $h$ too)?

Surjective. In order to show that $g$ is surjective, we take an arbitrary $(u,v) \in B$, and define $(x,y) = h(u,v) \in A$. This is a preimage since $g(x,y) = g\bigl(h(u,v)\bigr) = (u,v)$, as desired.

Injective. In order to show that $g$ is injective, we assume that $g(x_1, y_1) = g(x_2, y_2) \in B$ for some $(x_1, y_1), (x_2, y_2) \in A$. By applying the inverse $h$, we deduce equality of the preimages: $$ (x_1, y_1) = h\bigl(g(x_1, y_1)\bigr) = h\bigl(g(x_2, y_2)\bigr) = (x_2, y_2). $$