Inverse of multivariable mappings

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For a multivariable mapping to have an inverse, it must be both one-to-one and onto, meaning every output corresponds to a unique input and vice versa. This condition implies that mappings from higher-dimensional spaces to lower-dimensional ones cannot be invertible, as they cannot maintain a one-to-one correspondence. Specifically, a function from R^n to R^m, where n does not equal m, cannot be invertible due to dimensionality constraints. Additionally, when dealing with n independent variables, at least n equations are necessary to define a unique state. Therefore, the discussion emphasizes the importance of dimensionality in determining the invertibility of multivariable mappings.
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If there was a function f(x,y) = z...would the inverse be defined under injection (one-to-one correspondence) only if this condition was held?: Every z corresponds to only one (x,y) pair.

If this was the case...then wouldn't that imply that a good number of 2-D surfaces cannot have an inverse mapping?
Furthermore...what multivariable linear mapping would have an inverse (its not too hard to imagine multiple different orders of variables for which the same value is designated)?

Edit: my question is more or less this: What are the necessary conditions for an inverse for a multivariable mapping? If one-to-one correspondence is one of them, then what is the definition of one-to-one correspondence for multivariable mappings?
 
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Good question: the following ideas may help.




Let g: A -> B be a function from a set A into a set B.

Definitions of one-to-one and onto:

g is one-to-one iff for every a1, a2 contained in A, g(a1) = g(a2) implies that a1 = a2.

g is onto iff for every b contained in B, there exists at least one a contained in A such that
g(a) = b.

Let X, and Y be sets, and f:X ->Y be a function from X to Y. Then f has an inverse iff f is one-to-one and onto.



It turns out that for Euclidean spaces R^n, n >=1 n is the dimension of the space (n = 2, are two dimensional planes, for n = 3, you are looking at three dimensional flat space), like the ones you mentioned below, there exists a one-to-one and onto mapping between the spaces iff the spaces have the same dimension. Your mapping that you are looking at is a mapping from a two dimensional Euclidean space to a one dimensional Euclidean space, and therefore can not be one-to-one, and so can not be invertible.

The proof to the theorems above can be found readily via a quick google search, or probably somewhere on this website: it is a really good website for mathematics and physics, and other subjects in the sciences :)


Also, another way to think about this particular kind of problem, that may be quite intuitive:

if you have a function from the complex numbers into a vector space of the form
w = f(x1, x2, ..., xn), where where x1, x2, ... xn are independant variables are allowed to vary over a set of real or complex values in some subset A of the complex numbers, then
f(A) can have dimension at most equal to n. Functions of n independant variables map into spaces of dimension less than or equal to n.

Another general principle that usually holds: if you have n unknown parameter values in a system, you need at least n equations, conditions,...etc., to specify a unique state of the system.


Not everything above was stated precisely, they are just ways of thinking about the ideas: which ideas apply depend on the particulars of the concepts you are considering. Hope this helps.
 
Thanks. I thought this particular thread would die out, but a reply came (a good one). But to be clear, is what you were stating this?: Any mapping from Rn to Rx, x not equal to n, is not invertible.
 
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