Is $f: S^1 \rightarrow \Bbb{R}$ an embedding, submersion, or immersion? Defined via $f(x,y)=y$.

Question

Let

$$f: S^1 \rightarrow \Bbb{R}$$

be a smooth map given via

$$(x,y) \mapsto y$$

I can't it being surjective as large elements of $\Bbb{R}$ don't get "hit" and it cannot be $1-1$ as $(-1,0),(1,0)$ both map to $0$. My initial thought was its an immersion.

Answer 1

All of these questions are answered by looking at the differential $Df_{(0,1)}\colon T_{(0,1)}S^1\to T_1\mathbb R$.

Note that since $\dim(S^1)=\dim(\mathbb R)$, the dimension of their tangent spaces match, so a map $S^1\to\mathbb R$ being an immersion is equivalent to it being a submersion.

Answer 2

If you take the chart $\phi : (x,y)\to x$ on the up half of the circle, then $f\circ \phi ^{-1} = \sqrt{1-x^2}$ has its derivative equal zero at 0. What means that $f$ has a critical point on the point $(0,1)$.