Let $f\colon \mathbb{R}^2\rightarrow\mathbb{R}^3,\ f(u,v)=\begin{pmatrix} u \\ v^3(1+v^3) +\sinh(u)\\ v^3(1-v^3)-\sinh(u)\end{pmatrix}$
I want to prove that $f(\mathbb{R}^2)$ is an (embedded) submanifold of $\mathbb{R}^3$.
From the book I am reading (Amann and Escher - "Analysis II") I was endowed with the following theorem: Suppose $X$ is open in $\mathbb{R}^m$ and $f\in C^q(X,\mathbb{R}^n)$ is an embedding. Then $f(X)$ is an m-dimensional $C^q$ submanifold of $\mathbb{R}^n$.
An embedding is a immersion $f$ such that $f$ is topological onto its image $f(X)$, meaning $f^{-1}\in C^1(f(X))$.
A $C^1$-map is called immersion if its differential is injective for every element on its domain.
My attempt:
$f\in C^1(\mathbb{R}^2)$ since all of its coordinate functions $f_1,f_2,f_3$ are $C^1(\mathbb{R}^2)$. I now know that the differential $\partial f(x)$ is given by the Jacobi-Matrix
$[\partial f(x)] =\begin{pmatrix} 1 & 0 \\\ \cosh(u) & 3v^2+6v^5 \\\ -\cosh(u) & 3v^2-6v^5 \end{pmatrix}$.
I now need to find a $2\times 2$ invertible submatrix to prove that $[\partial f(x)]$ is injective, hence $f$ immersion.
I picked the last to columns and found the determinant to be $6v^2\cosh(u)$ which should not equal zero. Since $\cosh(x) >0, \ x\in\mathbb{R}$ this is granted for $v\ne 0$.
Now I want to check if $f^{-1} \in C^1(f(\mathbb{R}^2))$. I therefore set $f(u,v)=(x_1,x_2,x_3)$ and find that $x_1=u, \sqrt[3]{1/2(x_2+x_3)}=v$ which is problematic, because $v$ may not be defined , beacuse $x_2+x_3$ can be negative. (Choose $v<0$, we then cannot get a preimage, because the root is undefined)
Is my solution so far correct? What should I do with the point at which the determinant is not non-zero? How can I find an inverse of this parametrization?
"I now need to find a $2\times 2$ invertible submatrix to prove that $[\partial f(x)]$ is injective, hence $f$ immersion."
Actually you just need to check that the differential of $f$ has maximal rank at every point, namely, $2$. You do that by noting that the columns of $\partial f$ are linearly indpendent at every point $(u,v)$, which is obvious by inspection ($1$ is not proportional to zero - so the first column can never be a multiple of the second).