Prove $M$(Helix) is a sub-manifold of dimension $1$ of $\mathbb{R}^3$ 
Let $\alpha,\beta\in\mathbb{R}^+$. Consider the circular Helix  $M\subset\mathbb{R}^3$ the image of $\psi:[0,2\pi]\to\mathbb{R}^3,t\to(\alpha\cos t,\alpha\sin t,\beta t)$.
Prove $M$ is a sub-manifold of dimension $1$ of $\mathbb{R}^3$.

I have no ideia how to prove this. I have already proven $\psi$ is an injective immersion. I think I need to find a map $F:M\to\mathbb{R}$, which I happen to found $F:M\to\mathbb{R},(\alpha\cos t,\alpha\sin t,\beta t)\to(\alpha^2+\beta t^2)$. However this would imply the dimension to be 3-1=2, which is wrong.
Question:
How can I solve the question? Am I proceeding well?
 A: As defined, the Helix $M = \psi([0, 2\pi])$ is not a submanifold of $\Bbb R^3$, because it is not a topological manifold. Indeed, $\psi$ is an homeomorphism from $[0,2\pi]$ onto its image $M$, because it is continuous, injective, and goes from a compact space to a Hausdorff space. However, $[0, 2\pi]$ is not a topological manifold (it's a manifold with boundary), so $M$ cannot be a topological manifold.
Maybe you want to take the open Helix $M = \psi((0, 2\pi))$, $\psi : (0, 2\pi) \rightarrow M$.
The easiest way is not to work with submersions but with immersions. Indeed, using the definition of a submanifold for one direction and constant rank theorem for the other, one can prove the following :

$M \subset \Bbb R^n$ is a submanifold of dimension $m \leq n$ if and only if for all $p \in M$, there exists an open set $V \subset R^m$, an open set $W$, $p \in W \subset \Bbb R^n$, and an injective immersion $\phi : V \subset \Bbb R^m \rightarrow W \subset \Bbb R^n$ such that $\phi(V) = W \cap M$ and $\phi$ is an homeomorphism from $V \rightarrow W \cap M$.

In your case, set $M$ to be the open Helix, $m = 1$, $V = (0, 2\pi)$, $n = 3$, $W = \Bbb R^3$, $\phi = \psi$ (here we use the same sets $V$, $W$, and the same function $\phi$ for all $p \in M$).
