let $f : \mathbb{R} \to \mathbb{R^3}$ defined by $f(t) = (\cos t,\sin t,t)$
first as a hint it was suggested to show that $f : \mathbb{R} \to f(\mathbb{R})$ is a homeomorphism. (I have no idea how this will be helpful afterwards)
so clearly when restrained to $f(\mathbb{R})$ the map is surjective.
also $\text{dim} (\mathbb{R}) = 1 = \text{dim}(\text{Ker}f) + \text{rk} f$
because $f$ is surjective $ \iff \text{rk} f = 1$ then it follows that $\text{dim}(\text{Ker}f) = 0 \iff $ f is injective
the inverse $f^{-1}(t) = (\arccos t, \arcsin t, t)$ is continuous therefore $f$ is a homeomorphism, right ?
now how to find $T_{f(\mathbb{R})}(x_0)$ ? I know that $f(\mathbb{R})$ is an helix ?
I tried to write $f(\mathbb{R})$ as $\{(x,y,z) \;|\; x^2+y^2 = 1, z = \frac 12(\arccos x + \arcsin y)\}$
then considered the mapping $g(x,y,z) = (x^2+y^2 - 1, z - \frac 12(\arccos x + \arcsin y))$
$f(\mathbb{R}) = g^{-1}({0})$
$dg(x,y,z) = \begin{bmatrix} 2x & 2y & 0 \\ \frac {1}{2\sqrt{1-x^2}} &-\frac {1}{2\sqrt{1-y^2}} & 1\\ \end{bmatrix}$
which has rank $2$ whenever $x,y \neq 0 $
I hoped that it would be a submersion so I can construct the tangent space just from the Kernel but it wasn't.
Any help, hints will be greatly appreciated.
It makes no sense the use apply the rank-nullity theorem to a non-linear map.
The tangent space to $f(\mathbb{R})$ at some point $f(t_0)$ is the line$$\left\{f(t_0)+\lambda f'(t_0)\,\middle|\,\lambda\in\mathbb{R}\right\}=\left\{\left(\cos t_0,\sin t_0,t_0\right)+\lambda\left(-\sin t_0,\cos t_0,1\right)\,\middle|\,\lambda\in\mathbb{R}\right\}.$$