According to the inverse function theorem, a continuous, differentiable function from $\mathbb{R} ^n \mapsto \mathbb{R}^n$ is invertible if the determinant of the mapping function's Jacobian matrix is non-zero.
What I would like to clarify is if the associated Jacobian matrix can have an entry equal to zero.
For example, if we consider the map
$$\begin{bmatrix} x\\ y \end{bmatrix} \mapsto \begin{bmatrix} 5x\\ x + y \end{bmatrix} $$
The Jacobian has a non-zero determinant for all x, y (det = 5), but it appears to me that an inverse function does not exist because an inverse map from x prime, x prime being the point in the image that the point x in the preimage is mapped to, will be satisfied by all y co-ordinates and therefore is not a function. It appears that the (1, 2) entry of the Jacobian matrix being equal to zero is the problem.
In the 1D case, the requirement that the neighbourhood of a point must have non-zero derivative to be locally invertible prevents such a situation from occurring, so I would expect the same condition to feature with the condition that the Jacobian determinant is non-zero for the multivariable scenario, but I am unable to find any commentary about my problem.
Also, if I could squeeze in another related question, I am confused about why the theorem isn't an "if and only if", because my intuition suggests that the Jacobian det must be non-zero when defined for the function to be invertible, and if it isn't defined, then the function maps to a different dimension space but then you will always be able to find multiple co-ordinates satisfying a point and therefore this map is not injective and thus cannot be invertible anyway. For example, consider $\mathbb{R}^2 \mapsto \mathbb{R}$, so that f(x,y) = z, then any plane where z is equal to a constant would intersect at multiple points (a line). I would really appreciate if someone could formalise this, mostly just want clarification if a mapping to a different dimension space can ever be injective.
I would like to add that this is my first post so thank you very much for your time and attention, and I apologise for any amateur errors.