Considering the system of equations:

$x^2\cos(xy)=a$ ; $e^y=b$

Show that if $V$ is a significantly small neighbourhood of $(1,1)$, the system above has at least two solutions for each $(a,b)$ $\in$ $V$.

Firstly, I know that for $f(x,y)=(x^2\cos(xy),e^y)$ $\in$ $C^1$, $f(1,0)=(1,1)=f(-1,0)$ and that $\det(J_f(1,0))=2$ and $\det(J_f(1,0))=-2$ so by the Inverse Function Theorem, there are open sets $U,V$ that contain respectively $(-1,0)$ and $(1,0)$ such that $h(a,b)=(x,y)$ can be solved uniquely considering $(x,y) \in U$ and $(a,b) \in f(U)$ and also can be solved uniquely considering $(x,y) \in V$ and $(a,b) \in f(V)$. Now if we consider $A=f(U) \cap F(V)$, we have that for each $(a,b) \in A$, there will be at least two solutions one lying in $U$ and other in $V$ for $f(x,y)=(a,b)$. Could this procedure be considered?



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