# Bijection $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$

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 $$f(x,y) = \phi(x) + \psi(y)$$.

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 $$f(x)=F(x,0)$$, $$g(x)=F(0,x)$$.