Consider the following system of equations in $u,v,s,t$: $$(uv)^4+(u+s)^3+t=0; \\ \sin(uv)+e^v+t^2-1=0.$$ Prove that near the origin $0\in \mathbb R^4$ its solutions form a graph of a $C^1$ function $G:\mathbb R^2\to \mathbb R^2$. Clearly indicate the dependent and independent variables.
I consider the map $f:\mathbb R^{2+2}\to \mathbb R^2$ given by $$(u,v,s,t)\mapsto((uv)^4+(u+s)^3+t,\sin(uv)+e^v+t^2-1).$$ Note $f(0)=0$. The implicit function theorem says that if the Jacobian of $f$ w.r.t. $s,t$ is non-zero at the origin, then there is a neighborhood $U\subset \mathbb R^2$ and a $C^1$ function $G: U\to \mathbb R^2$ such that for all $(u,v)\in U$ we have $f(u,v, G(u,v))=0$ (note that $G(u,v)$ lives in $\mathbb R^2$, so the expression $f(u,v, G(u,v))$ is well-defined), and so the solutions are the graph of the function $G: U\to \mathbb R^2$.
But there are two problems:
The function $G$ has $U$ as a domain, not the whole $\mathbb R^2$.
The Jacobian w.r.t. $(s,t)$ at $(u,v,s,t)$ is $3(u+s)^2\cdot 2t$, which is zero at the origin.
What am I doing wrong?