Let f: $R^{2}$ -> $R^{2}$ be given by: f(x,y):= \begin{pmatrix} x^{4} + y^{4} \\ 4xy\end{pmatrix}
- Show that f is locally invertible in any point of the line in $R^{2}$ defined by the equation 2y = x .
- Is f globally invertible? Show working
How would one solve such questions. Can someone please give a detailed working to clear up my concept.
Local Invertibility
Local invertibility refers to whether the closest affine approximation to $f$ at a given point $(x,y)$ is invertible. The closest affine approximation to $f$ at a point $(x,y)$ is given by $f(x,y)$ plus the derivative of $f$, otherwise known as the Jacobian in this case.
Denote $f(x,y) = \left(\begin{matrix} f_1(x,y) \\ f_2(x,y) \end{matrix}\right)$. Then the Jacobian of $f$ at a point $(a,b)$ is a linear map $\mathbb R^2 \rightarrow \mathbb R^2$ given by the matrix
$$J(a,b) = \left.\left( \begin{matrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_2}{\partial x} \\ \frac{\partial f_1}{\partial y} & \frac{\partial f_2}{\partial y} \end{matrix} \right)\right|_{x=a,y=b}.$$
In this case, it's a straightforward computation to yield
$$J(x,y) = \left( \begin{matrix} 4x^3 & 4y^3 \\ 4y & 4x\end{matrix} \right).$$ The question asks whether $f$ is locally invertible along a particular line in its domain defined by $x=2y$. This is equivalent to the Jacobian of $f$ along this line being invertible. How do we test whether a linear map $\mathbb R^2 \rightarrow \mathbb R^2$ is invertible? Just test its determinant!
Substituting $x=2y$ into the Jacobian: $$J(2y,y) = \left( \begin{matrix} 32y^3 & 4y^3 \\ 4y & 8y\end{matrix} \right).$$ Calculating its determinant: $$\begin{align}\det J(2y,y) &= (32y^3)(8y) - (4y^3)(4y) \\ &=y \left( (32y^2)(8) - (4y^2)(4)\right) \end{align}$$ ...and we stop right there, because we can see that at $y=0$, this determinant is $0$. The map $f$ is in fact not locally invertible along the line of the domain $(2y,y)$.
Global Invertibility
It is easy to see that $f$ is not globally invertible, since it is not an injective map. It is symmetric in $x$ and $y$, meaning that we can swap these parameters and retain the same value of $f$. So pick any non-zero $x$ or $y$ and we find $$f(x,y) = f(y,x)$$ ...meaning that $f$ assigns the same value to two distinct points in its domain.