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I want "something" ("something" because maybe it is not really a mathematical function, called F in the above image) that can describe what is shown on the image. A given value from a domain Xi can be mapped to many values of domain X{i+1}, and for two values $x \neq x'$, then $F(x) \neq F(x')$

It also should be invertible, i.e. $F^{-1}$ exists.

Is it possible to have this ? If not, how can I have an alternative for this with something possible in the mathematics ? Maybe using a function where the elements of the codomain are vectors ? ...

EDIT: I'm not just seeking a name for that. Let's call it a "relation". I want to define a relation F that allow me to generate a set of numbers $y1, y2, y3$ given a number x (that is, F(x) = {y1, y2, y3}), and I want it to be invertible, that is $F^{-1}(y1) = x$, $F^{-1}(y2) = x$ and $F^{-1}(y3) = x$.

This numbers (elements of each domain $X_i$) may be natural, real, complex, or whatever ...

Also, note that the domains $X_i$ does not overlap, that is elements are unique: for any given element x from $X_i$, there is no element y in any other $X_j$ such that x=y.

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    Comparing with "Deduce a unique ...", it seems that (working with $\mathbb N_0$) letting $F(n)=\{kN+1,\ldots, k(N+1)\}$ with $F^{-1}(n)=\lfloor \frac{n-1}k\rfloor$ is one example of such a beast (for any fixed $k\in \mathbb N$).2012-10-03

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There is a concept called multivalued functions. A multivalued function from $X$ to $Y$ is often denoted by $f:X\leadsto Y$, and is really a map $f:X\to {\cal P}(Y)$ from $X$ to the power set of $Y$. You can of course add various constraints on $f$. I haven't seen the requirement of disjoint sets $f(x)$ and $f(x')$ for $x\neq x'$ that you desire, but then I'm more interested in cases where the sets $f(x)$ vary continuously with respect to $x$.

In fact, you can do serious analysis with multivalued functions, as described in Set-Valued Analysis by Aubin and Frankowska.

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    Is it possible to have an inverse function for the multivalued function, so that if the multivalued function is F(x) = {y,z,k} then $F^{-1}(y) = x, F^{-1}(z) = x, F^{-1}(k) = x$ ? For example if we take the "root square" function (or multivalued-function?) $\sqrt{4} = \{-2, 2\}$, then there is a convenient inverse function which is "squaring": $({-2})^2 = 4$, $2^2 = 4$.2012-10-03
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Functions can be viewed as sets of ordered pairs - loosely, $f=\{(x,f(x)):x\in \mathcal{D}\}$ for some domain $\mathcal{D}$. The requirement that defines a function is that it gives a "good recipe" - that is, you plug an $x_0\in \mathcal{D}$ into it, it gives one and only one value $f(x_0)$, as to be non-ambiguous. In the ordered pairs picture of functions, this means that $f$ only contains one ordered pair with $x_0$ in the first position.

You wish to take away that requirement, so the object you want to make is not a function. But that's okay - we can still make things like that by viewing $f$ as a set of ordered pairs. Since you want $f^{-1}$ to exist, insist that any value $y$ in the codomain appears only once in the ordered pairs (just as we would require for a function to be one-to-one). In mathematical language $f$ is now simply a relation, not a function, but it does everything you'd like it to do.

Note that this works no matter which sets are your domain and codomain - it doesn't matter whether you're working with reals, complexes, or even numbers at all.

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    @redfiloux I second that; branch cuts are super neat.2012-10-04