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Function is bijective when it is injective and surjective. Function is injective if $$(\forall x_1,x_2 \in A)f(x_1)=f(x_2)\Rightarrow x_1=x_2$$ and surjective if $$(\forall y \in B)(\exists x \in A)y=f(x)$$ Is it possible to transform a function, so that it is given in explicit form? I don't know how to manipulate with function in this form ($f(x,y)=(2x-y,x-4y)$).

Thanks for replies.

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  • $\begingroup$ Try to solve $(z,w)=(2x-y,x-4y)$ for $(x,y)$. $\endgroup$ – Christoph Jun 26 '15 at 11:12
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You can use matrices. your function $f$ can be written as

$$ A=\left( \begin{array}{rl} 2& -1 \\ 1 & -4 \end{array}\right). $$ Here $f(x,y)=A\binom{x}{y}$.

By finding $A^{-1}$ you can describe $f^{-1}$.

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Surjectivity: take a target $(a,b)$

Tou have $2x-y=a$ and $x-4y=b$, which is a system that solves like that:

$x=\dfrac{4a-b}{7}$ and $y=\dfrac{a-2b}{7}$. Thus every target can be reached, and $f$ is surjective.

Injectivity is kind of easier even.

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Let $(z_{1},z_{2}) \in \mathbb{R}^{2}$. Then some $x,y \in \mathbb{R}$ are such that $z_{1} = 2x-y$ and $z_{2} = x-4y$ if and only if $$ x = \frac{4z_{1}- z_{2}}{7}, y = \frac{z_{1} - 2z_{2}}{7}, $$ so $f$ is bijective and $$f^{-1}: (z_{1},z_{2}) \mapsto \bigg( \frac{4z_{1}- z_{2}}{7}, \frac{z_{1} - 2z_{2}}{7} \bigg)$$ on $\mathbb{R}^{2}$ in $\mathbb{R}^{2}$.

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  • $\begingroup$ Is it possible to find $f^{-1}$ explicitly, without using matrices? $\endgroup$ – user300045 Jun 26 '15 at 11:22
  • $\begingroup$ Sorry? I did not use any matrix theory. What I showed is that for every $(z_{1},z_{2}) \in \mathbb{R}^{2}$ there is exactly one $(x,y) \in \mathbb{R}^{2}$ such that $f(x,y) = (z_{1},z_{2})$, so the bijectivity follows. But that also gives the inverse function $f^{-1}$. $\endgroup$ – Megadeth Jun 26 '15 at 11:30

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