 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2013/f/1/c/iii/solution

G

C^{1}

B

D G (x) = (x_{3} + 3)^{- 1} 00 (x_{3} + 3)^{- 1} 10 - (x_{1} + x_{2}) (x_{3} + 3)^{- 2} 01 .

At the origin

G (0) = 0

and

det D G (0) = 1/3 \neq = 0

. The inverse function theorem therefore supplies the required open sets

V_{1} \subset B

and

V_{2} \subset R^{3}

, both containing zero, with

G : V_{1} \to V_{2}

a diffeomorphism. The explicit inverse formula on its image,

G^{- 1} (y_{1}, y_{2}, y_{3}) = (y_{1} (y_{3} + 3) - y_{2}, y_{2}, y_{3}),

also checks the local conclusion directly.

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