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I need a bijection, such that: $f(x,y) = \phi(x) + \psi(y)$ and $f(x,y) = g(\phi(x) + \psi(y))$. Second one is easy, I think. If we make first one, we can consider $g$ as identical map. I have seen similar topics on stack, but I don't like those answers. I think it is possible to construct a bijection, as Cantor did, proving that $[0,1] \times [0,1]$ ~ $[0,1]$, because I can build a bijection $h: [0,1] \to \mathbb{R}$. Thus we have a equivalent problem, with segment, i.e build a bijection $f: [0,1] \times [0,1] \to [0,1]$, such that $f(x,y) = \phi(x) + \psi(y)$.

Thank you in advance!

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For the first one, you can consider $R$ as a $Q$ vector space of infinite dimension, and choose a base, say $(e_{\alpha})_{\alpha \in A}$. Now choose any bijection $A\times A \to A$. this bijection yields a bijection between the bases of $R\times R$ and $R$ therefore a $Q$-linear isomorphism, say $F: R\times R\to R$, and $F(x,y)=F(x,0)+F(0,x)$. Set $f(x)=F(x,0)$, $g(x)=F(0,x)$.

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