 Solution

ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2021/4/c/i/solution

Write

a = \partial_{1} f_{1} = \partial_{2} f_{2}

and

b = \partial_{2} f_{1} = - \partial_{1} f_{2}

D F_{x} = (a - b b a), det D F_{x} = a^{2} + b^{2} .

Consequently

D F_{x} is invertible exactly when (\partial_{1} f_{1} (x), \partial_{2} f_{1} (x)) \neq = (0, 0) .

Equivalently, it is singular exactly when all four partial derivatives vanish. The sum of squares is over

R

, so no cancellation is possible.

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