Part CCSIR NET June 2024immersion-is-not-submersion

Immersion is not submersion

Let $f : \mathbb{R}^{2} \to \mathbb{R}^{3}$ be a differentiable function such that (Df)(0,0) has rank 2. Write $f = (f_{1}, f_{2}, f_{3})$ . Which of the following statements are necessarily true?

A.f is injective in a neighbourhood of (0, 0)✓ Correct
B.There exists an open neighbourhood U of (0, 0) in $\mathbb{R}^{2}$ such that $f_{3}$ is a function of $f_{1}$ and $f_{2}$
C.f maps an open neighbourhood of (0, 0) in $\mathbb{R}^{2}$ onto an open subset of $\mathbb{R}^{3}$
D.(0, 0) is an isolated point of f⁻ $^{1}($ {f(0,0)})✓ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests converse assumed.

See pricing

50 are analysed free — try those first.

The trap it tests

Converse assumed

The theorem runs one way. You used it in the other.

Drill statements like this

Immersion is not submersion

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