Is there any polynomial $f(x,y)\in{\mathbb Q}[x,y]{}\ $ such that $f:\mathbb{Q}\times\mathbb{Q} \rightarrow\mathbb{Q}$ is a bijection?

Let us consider a simpler case first. Let $Q_1=Q\cap [0,1]$. Now let us assume $f:Q_1\times Q_1\to Q_1$ is a uniformly continuous bijection. Then, according to Rudin (I believe it is an exercise), there is unique continuous extension, $g$, such that: 1) $f=g$ on $Q_1\times Q_1$. 2) $g$ is uniformly continuous on $\overline{Q_1\times Q_1}=[0,1]^2$. 3) Namely, $g(x_0)=\lim_{x\to x_0} f(x)$. Since the image of a connected, compact set is connected and compact, then $Im(g)=[0,1]$. But, this is impossible because if we consider three distinct rational points $a,b,c$ in $[0,1]^2$, then $g$ restricted to the connected set $[0,1]^2${$a,b,c$} is still continuous, but the image will not be connected since g is bijective on $Q_1\times Q_1$. I think the above case now follows. That is if we consider $f:Q\times Q\to Q$ and $f$ is a polynomial, then $f$ restricted to $Q_1\times Q_1$ would could be a uniformly continuous bijection. However, we won't know what the image is. But this doesn't matter, since when we extend to $g$ the continuous image of a connected compact set is connected and compact. But the only connected compact subsets of $R$ are bounded closed intervals. So, the image would be some $[a,b]$ and would get another contradiction. 


Edit[ The following is wrong ~ see comments] I don't think so. Suppose $f$ to be surjective. Let $x\mapsto 0$ and $y\mapsto 1$. Now consider two distinct paths $\gamma,\eta:[0,1]\to\mathbb Q\times\mathbb Q$ from $x$ to $y$. Since $f$ is continuous it maps these paths surjectively onto $[0,1]$ (more exactly $[0,1]\subset f\gamma([0,1])$ and $[0,1]\subset f\eta([0,1])$). Thus, $f$ cannot be injective. 


protected by Andrés Caicedo Dec 5 '13 at 14:23
Thank you for your interest in this question.
Because it has attracted lowquality answers, posting an answer now requires 10 reputation on this site.
Would you like to answer one of these unanswered questions instead?