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Is the inverse of a bijective binary algebraic function also an algebraic function?

Take for example:

$x_1,x_2\in\mathbb{R}$

$F(x_1,x_2)=x_1+x_2$

$y=F(x_1,x_2)$

$y=x_1+x_2$

$x_1=y-x_2$, $x_2=y-x_1$

$(x_1,x_2)=(y-x_2,y-x_1)$

For determining the arguments $(x_1,x_2)$ of $F$, only algebraic operations are needed.

We see: The inverse $F^{-1}$ of $F$ is a 2-valued function. But what is a 2-valued algebraic function?

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    No such thing as a 2-valued function. This function is also not bijective. – 2017-02-05
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    $F$ should be bijective. Take e.g. $x_1=x$, $x_2=e^x$, $x\in \mathbb{R}$. The domain of $F$ is $\{ (x_1,x_2) \in \mathbb{R}^2 | x_1=x, x_2=e^x\}$. $F$ is binary and bijective. Its inverse is 2-valued. But is it algebraic? – 2017-02-05

1 Answers 1

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The inverse relation of a bijective binary function is a 2-valued function, a multivalued function. If this inverse relation should be an algebraic function, it has to be a function. But a function cannot be multivalued. A 2-valued function is no function. "2-valued function" and "multivalued function" are misnomers because a function can be only one-to-one or many-to-one. For treating multivalued functions as functions, you have to treat them e.g. as set-valued functions or as tuple-valued functions.

An algebraic function $f$ with $y=f(x_1,x_2,...,x_n)$ is by definition a root of a polynomial equation (the defining algebraic equation of $f$)

$$p_0(x_1,...,x_n)+p_1(x_1,...,x_n)y+p_2(x_1,...,x_n)y^2+...+p_n(x_1,...,x_n)y^n=0$$

where the $p_i$ ($\forall i=0,1,2,...,n$) are polynomials in $x_1,x_2,...,x_n$. If you want to allow $f$ to be set-valued or tuple-valued, you have to allow powers of sets (respectively powers of tuples) for $y^2,y^3,...,y^n$ in the defining algebraic equation of $f$.