Given the question as below:(apology, I don't know how to format properly yet)
I am aware that a map $$f: X \to Y $$
is called a local diffeomorphism if
1)$F$ is onto and
2)for every $x \in X$, there exist open neighbourhoods $U$ of $x \in X$ and $V$ of $f(x) \in Y$ such that $$f|_U: U\to V $$ is a diffeomorphism
So my attempt is as followed:
$f(t)= (\cos(t), \sin(t))$
If I consider $f:[0,1] \to S $
i.e., $$f(t) = e^{2{\pi}it}$$
then clearly, $f$ is a differentiable, continuous and onto function in that range.
Hence, $f$ is smooth.(1 is proven)
I am a bit stuck on how to prove 2. My attempt is as follows:
Suppose there exist $X$- a non empty subset of $\mathbb{R}$
Let
$$f: X\to \mathbb S^1$$
Since $f$ is smooth, meaning for every $x \in X$ there exist an open neighborhood $U$ of $x$ in $\mathbb{R}$ and a smooth map
$$F:U \to \mathbb S^1$$
such that $F(y)=f(y)$ for every $y \in X \cap U$
From then on, I was thinking of using the definition of smooth manifold to create an open nbh $U$ in $X$ and a diffeomorphism to optain a smooth chart, hence $X$ is a locally diffeomorphism.
Is my understanding correct? I'd appreciate for all the help and guidance.