Measure of invertibility of functions I've been struggling to find a method of describing 'how' invertible a function is. For instance, $f(x)=x$ should be totally invertible, while $f(x)=x^2$ should be half invertible (or something like this). Don't know if the inverse of the norm of the Jacobian's determinant is going to be of any help here?
I kind of view it as a ratio between the'size' of image and the 'size' of the domain.
I feel that such a notion simply must exist and I'm just not looking in the right places.
 A: One way to measure the failure of a (linear) function to be invertible is the kernel of it.
This is defined as (for $f:X\to Y$):
$$\ker f = \{x\in X\mid f(x) = 0\}$$
Clearly, for a (linear) function, $\ker f = \{0\}$ implies $f$ is invertible (in fact, they're equivalent statements).
We can generalize the notion of kernel by looking at everything that gets sent not to $0$, but to arbitrary $k\in Y$.
This set would look like:
$$\{x\in X\mid f(x) = k\}$$
This is known as the pre-image, and is denoted $f^{-1}(k)$.  Note that this isn't the inverse, as $f^{-1}(k)$ is a set.
The pre-image at a particular point having multiple elements in it measures precisely what you want --- for $f(x) = x^2$, we have that:
$$f^{-1}(k) = \begin{cases} \{\sqrt{k},-\sqrt{k}\} & k \neq 0 \\ \{0\} & k = 0\end{cases}$$
We then have that:
$$|f^{-1}(k)| = \begin{cases} 2 & x \neq 0 \\ 1 & x = 0\end{cases}$$
So, the size of this preimage not being $1$ tells us the function isn't invertible at that point.  The above is all standard stuff.
Using this, you can probably measure "how much of the co-domain isn't invertible" in some way just by looking at what proportion of the codomain has $|f^{-1}(y)| = 1$ (or $|f^{-1}(y)| = 0$ for that matter).  I'm unfamiliar if there are any standard techniques along these lines.
