Question

Let \( f: \mathbb{R}^2 \to \mathbb{R}^2 \) be defined as
\[ f(x, y) = (e^x \cos y,\ e^x \sin y) \]
Which one of the following is TRUE?

• A. \( f \) is one-to-one.
• B. The Jacobian of \( f \) is negative.
• C. \( f \) is locally invertible.
• D. \( f \) is invertible on \( \mathbb{R}^2 \).

Hint:

Compute the Jacobian determinant of f, then compare local invertibility with global injectivity using the periodicity of sine and cosine in y.

Answer 1

The Correct Option is C

Step 1: Write the map in terms of the Jacobian.
The map is \( f(x,y) = (e^x \cos y,\ e^x \sin y) \). Call the two component functions \( u(x,y) = e^x \cos y \) and \( v(x,y) = e^x \sin y \). To check invertibility we first find the Jacobian matrix of partial derivatives.

Step 2: Compute the Jacobian matrix and its determinant.
\[ u_x = e^x \cos y, \quad u_y = -e^x \sin y, \quad v_x = e^x \sin y, \quad v_y = e^x \cos y \]
So the Jacobian determinant is
\[ J = u_x v_y - u_y v_x = e^{2x}\cos^2 y + e^{2x}\sin^2 y = e^{2x} \]
Since \( e^{2x} > 0 \) for every real \( x \), the Jacobian is never zero and never negative. This already rules out option (B).

Step 3: Use the Inverse Function Theorem.
The Inverse Function Theorem says that if the Jacobian determinant of a continuously differentiable map is nonzero at a point, the map is locally invertible near that point, meaning it has a smooth local inverse on some small neighborhood.
Here \( J = e^{2x} \neq 0 \) at every point of \( \mathbb{R}^2 \), so \( f \) is locally invertible at every point. This makes option (C) true.

Step 4: Check global injectivity, which is why A and D fail.
Being locally invertible everywhere does not mean \( f \) is globally one-to-one. Notice that \( f(x, y) \) depends on \( y \) only through \( \cos y \) and \( \sin y \), and both repeat every \( 2\pi \).
So \( f(0, 0) = (e^0\cos 0,\ e^0 \sin 0) = (1, 0) \) and \( f(0, 2\pi) = (e^0 \cos 2\pi,\ e^0 \sin 2\pi) = (1, 0) \) as well.
Two different points, \( (0,0) \) and \( (0, 2\pi) \), give the same output, so \( f \) is not one-to-one. This rules out option (A). Since \( f \) fails to be injective on all of \( \mathbb{R}^2 \), it cannot be invertible on \( \mathbb{R}^2 \) either, which rules out option (D).

Final Answer:
The Jacobian of \( f \) is nonzero everywhere, so by the Inverse Function Theorem, \( f \) is locally invertible at every point of \( \mathbb{R}^2 \), even though it is not globally one-to-one. \[ \boxed{\text{(C) } f \text{ is locally invertible.}} \]