Not surjective function by using algebra tools

Question

Let $T$ be a transcendental basis and $E\subset T$ over $\Bbb Q$ and let $f\colon \Bbb R\to \Bbb Q\cdot E$ such that $f(0)=0.$ Put, $D:=\{x\colon f(x)\neq 0\}.$ Define a function $g\colon \Bbb R\to\Bbb R$ such that $g(x)=\frac{f(x)}{x}$ for each $x\in D$ and $g(x)=0$ otherwise.

$g$ can not be a surjective function?

My attempt . Notice that let $ x\in D$ there exists $e_x\in E$ such that $f(x)\in\Bbb Q\cdot e_x$ and a finite $S\subset T$ such that $\frac{1}{x}\in \overline{\Bbb Q(S)}$, where $\overline{\Bbb Q(S)}$ is the algebraic closure of $\Bbb Q(S).$ Since $g(x)=\frac{q\cdot e_x}{x}$ so we have $g(x)\in\overline{\Bbb Q(S\cup\{e_x\})}.$

Any idea how I can finish ?

Answer 1

It's certainly possible for $g$ to be surjective. Pick an enumeration $(x_\alpha)_{\alpha<\mathfrak{c}}$ of $\mathbb{R}\setminus\{0\}$. You can then try to define a sequence $(y_\alpha)_{\alpha<\mathfrak{c}}$ of distinct reals by transfinite recursion such that $x_\alpha y_\alpha\in E$ for each $\alpha$ (the idea being you want $g$ to map $y_\alpha$ to $x_\alpha$). As long as $|E|=\mathfrak{c}$ this will be possible, since at each step of the recursion there will exist some $e\in E$ such that $e/x_\alpha$ is not equal to any of the $y_\beta$ you have chosen already. If you then define $f$ such that $f(y_\alpha)=x_\alpha y_\alpha$ for each $\alpha$, then $g(y_\alpha)$ will be $x_\alpha$ so $g$ will be surjective.

Answer 2

$g$ can very easily be surjective.

Let $f$ be the function that sends $0\to0$, everything else to some $e\in E$.

Thus, $$g(x)=\begin{cases}0&x=0\\e\cdot x^{-1}&x\neq0\end{cases}$$For any $y\in\mathbb R\setminus\{0\}$, $g(ey^{-1})=e\cdot e^{-1}\cdot y=y$. Thus, $g$ is surjective since $g(0)=0$.