An injective map where each value is mapped to many others?

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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Relations, do you maybe just seek a name for that? I'd call it left-total left-unique relation. –  k.stm Oct 3 '12 at 21:30
@K.Stm. I'm not just seeking a name for that. Ok let's call it a "relation". I want to define a relation 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 $F^{-1}(y3) = x$. –  shn Oct 3 '12 at 21:37
I understand that you donot only want $F(x)\ne F(x')$ for $x\ne x'$ but rather $F(x)\cap F(x')=\emptyset$. Then this is a (possibly infinite) rooted tree with $X_i=$ the set of vertices at level $i$. The inverse map is the "father" map –  Hagen von Eitzen Oct 3 '12 at 21:40
Possible duplicates: Create a unique numbers‌​, Deduce a unique number from number –  Hagen von Eitzen Oct 3 '12 at 21:45
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 $k\in \mathbb N$). –  Hagen von Eitzen Oct 3 '12 at 21:51

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": $({-2})^2 = 4$, $2^2 = 4$. –  shn Oct 3 '12 at 22:45

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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By the way, I should add that a simple version of what you're looking for is the square root "function." Square rooting a number $a$ yields two distinct solutions $\pm \sqrt{a}$, yet the inverse function (squaring) gives you the same number back for each. –  Alexander Gruber Oct 3 '12 at 22:07
Concerning your comment: Why the square root function is actually called "function" since yields two distance values ? Isn't it a multivalued function ? Furthermore ot have a convenient inverse which gives back the same number for each ! Don't you have another similar example of function which yields to more than two distinct solutions ? –  shn Oct 3 '12 at 22:38
Concerning your answer: since I want $f^{-1}$ to exist, that means for me that it is useless to have "pairs" (x, f(x)), because we can anyway have the value of x by doing it on $f^{-1}(f(x))$, then f(x) is sufficient to find out x, which means that we do not need to have pairs of (x, f(x)). It is very clear. –  shn Oct 3 '12 at 22:42
@shn I put "function" in quotation marks because square root is not actually a function, per se, for exactly the reasons you've said. A lot of times people define a square root function on the reals by (arbitrarily) choosing the positive value. This gets a bit more complicated when dealing with complex numbers or $n$-th roots, but the concept remains the same - the formally make these into a function, you have to pick a single value to use every time. –  Alexander Gruber Oct 4 '12 at 1:03
You may have fun playing around with the complex logarithm en.wikipedia.org/wiki/Complex_logarithm ! –  Bogdan Oct 4 '12 at 2:10