How to prove this mapping is algebraic?

Solution 1:

To make a quick application of the standard inverse function theorem in this example, consider $F(u,v)=(f(u,v),g(u,v))=(u+v,1/uv)$. Near specific $(u_o,v_o)$, there will be a ("local") differentiable inverse function if $\det\pmatrix{f_u & g_u \cr f_v & g_v}\not=0$, where the subscripts are partial derivatives. Here this condition is $$ 0 \;\not=\; \det\pmatrix{ 1 & {-1\over u^2v} \cr 1 & {-1\over uv^2}} \;=\; {-1\over uv^2} - {-1\over u^2v} \;=\; {1\over u^2v^2}\cdot (v-u) $$ This is non-zero for $u\not=v$, and away from $u=0$ or $v=0$, so in that region there is a (local!) inverse function, by the theorem.

Yes, this guarantees existence without worrying about formulas.