Is the curve $S=\{(|\sin(t), \cos(t) \exp(t)) \in\mathbb{R}^2 \mid t \in(0,{3\pi\over4})\}$ a manifold in $\mathbb{R}^2$?

Question

Is the curve $S=\{(|\sin(t), \cos(t) \exp(t)) \in\mathbb{R}^2 \mid t \in(0,{3\pi\over4})\}$ a manifold in $\mathbb{R}^2$?

I feel it is a manifold, as when you look at the graph, it does not overlap itself at all, so S maps injectively. But everything I'm feeling is based purely on intuition, and I am unsure how to formally go about proving this either way.

Answer 1

Yes, it is an embedded submanifold. This is because the map $$ f(t) = (\sin(t),\cos(t)e^t) $$ satisfies the following sufficient conditions of an embedding:

1. $f$ is injective (as you pointed out)
2. $f$ is an immersion (since the tangent vector is never 0)
3. $f$ is a homeomorphism between $(0,3\pi/4)$ and the image curve