A special function

If we're transforming a line, f(x)=1, we could reparametrize this line in vector form as [x,f(x)] = [t, 0]. Then if we subject this to a one-to-one matrix transformation A[x,f(x)]+B, where A and B are matrices with constant entries, we can arrive at [x,f(x)]=[0,t]. Even though x → f(x) is no longer one-to-one, the resulting matrix is one-to-one. :|

The idea is that the injective function f(x) has an inverse so that f(x) = 1/f⁻¹(x). I don't see how this solves the problem, sadly.

 Spoiler

Alternatively, you could force f(x) = x^±i to return real numbers by having your domain be a set of complex numbers.

Well, infinity is not in ℝ The other idea of your works perfectly, though!

A simple function in a subset of ℝ is the function 2³¹ - 1 has mentioned for quite some time: f(x) = 1, f⁻¹(x) = 1 on the domain [1, 1]. It is unquestionably useless though.

Hmm... I just realized something. If we restrict the domain of f(x)=x^2 to two sections, then we can find "two" inverses:

For f(x)=x^2, x>0
f⁻¹(x)=sqrt(x)

For f(x)=x^2, x<0
f⁻¹(x)= -sqrt(x)

Thus, when we put the two together, we once again see the reflection over the line f(x)=x. Since both of them are true inverses, the put together "pseudo"-inverse still as a whole satisfies being the inverse of f(x). If we had a periodic function such as f(x)=sin(x), then we can just restrict it infinitely many times so that the domains are one-to-one everywhere. When we put together all the pieces, we once again end up with the reflection across the line f(x)=x.

Correct! Split the function at the local maximas and local minimas, as you'd then get a set of injective functions. Find the inverses, and voilà!

Noah