Well, briefly heres what I have been able to gather about relations and functions. Please do correct me if I go wrong somewhere.

A mathematical rule associating i.e. relating the elements of one set (lets call this set A) to the elements of another set (lets call this set B) is called a relation from set A to set B.

The relation may or may not relate all the elements of set A to elements of set B. Also every element in B may or may not be such that an element of A is related to it.

many-many relation:
If there is at least one element in A which is related to more than one elements in set B AND there is at least one element in B such that more than one elements in set A relate to it, then such a relation is called a ''many-many'' relation.

one-many relation:
If there is at least one element in A which is related to more than one elements in set B AND there is no element in B such that more than one elements in set A relate to it, then such a relation is called a ''one-many'' relation.

many-one relation:
If there is at least one element in B such that more than one elements in set A relate to it AND there no element in A such that more than one elements in set B are related to it, then such a relation is called a ''many-one'' relation.

one-one relation:
If there is no element in A which is related to more than one elements in set B AND there is no element in B such that more than one elements in set A relate to it, then such a relation is called a ''one-one'' relation.


A function (generally denoted by f(x)) is a special case of a relation which relates EVERY element of set A to not more than one element of set B. However every element of B need not be such that some element of A is related to it. Set A is called the domain of the function and set B is called the codomain.
Thus many-one and one-one relations MAY be (not necessarily are) functions but many-many and one-many relations can never be functions.
Those elements of the codomain, to which some element of the domain relate to, form a set called the range of the function.
A function for which range = codomain is called an onto function.
An inverse function exists only for a one-one onto function.

(I hope somebody does read all this )

I fully agree with that, pacman2003.

quote:Original post by Anonymous Poster
That still sounds weird to me. If there exists an inverse for f(x), I can't see how that condition could ever be met, as it seems to contradict the very definition of an inverse function.

Okay, in Euclidean space it would be hard to find such a function if we just thought about typical geometry. One obvious function that comes to mind though: y=Ez(p(z|x)) for stochastic variables x and z. y is not stochastic. Ez is the expectation taken w.r.t z. For nonlinear functions p(), the inverse of this function does not map y back to the original value(s) of x.

I hope this helps to highlight what I was saying above.

Cheers,

Timkin

[edited by - Timkin on January 29, 2004 7:18:43 PM]

quote:Original post by Timkin
I hope this helps to highlight what I was saying above.

Not really. In fact, it just sounds even weirder! I can''t see how bringing stochastic variables and nonlinear probability functions into the picture would make an inverse function not being an inverse function. I would be thrilled though if you would care to explain things a bit further, especially your definition of an inverse function.

Anyway, you say above that this is the ''typical'' way to describe a function not being ''onto'', yet I have not seen anything quite like it in any of my books or searching the internet. Rather, the typical way to put it seems to be that a function is not ''onto'' if the range differs from the codomain. To clarify your position, would you say that your definition is compatible with this view, and would you agree that in pacman2003''s example, f(x) is not ''onto'', for A=2?

quote:Original post by Anonymous Poster
I can't see how bringing stochastic variables and nonlinear probability functions into the picture would make an inverse function not being an inverse function.

I think you've misunderstood what I wrote. I'm not talking about an inverse function thats not an inverse. I'm talking about situations where x!=f-1(f(x)). For a function to be onto, it must be the case that x=f-1(f(x)). If this does not hold true, then the function cannot be onto.

quote:Original post by Anonymous Poster
I would be thrilled though if you would care to explain things a bit further, especially your definition of an inverse function.

My definition of an inverse function is the mathematical definition we all use. That is, take a function y=f(x). The inverse function of f, denoted f-1, is obtained by swapping y for x and x for y (giving x=f(y)) and then rearranging to make y the subject again. The final function is f-1. I'm sure we'd all agree on that, since its a standard definition.

quote:Original post by Anonymous Poster
Rather, the typical way to put it seems to be that a function is not 'onto' if the range differs from the codomain.

...and how would you describe that mathematically? As above, which is certainly the way that I have seen it presented, both during my education and in literature since. Of course, every book is different and its quite possible that our books differ in their explanation. I'm fairly certain though that we're talking about the same thing (excluding your belief that I'm talking about inverse functions that aren't inverse functions, which I'm not).

quote:Original post by Anonymous Poster
...would you agree that in pacman2003's example, f(x) is not 'onto', for A=2?

First, let's make it clear that there is obviously a problem with the function as presented. Clearly it does not map R ->R . In fact it maps R /{-0.25,1}->R /{-0.25}.

I may have made an error, however assuming I haven't, then the inverse function I get is:

f-1(x) = 3(x-1)+-5(x+1)         -------------            2(4x+1)   

The domain of this function excludes the point -0.25, so the domain of the inverse coincides with the range of the function, namely R /{-0.25}. According to your definition, that makes the function onto. However, is it the case that x=f-1(f(x))? In other words, does the range of the inverse coincide with the domain of the function? The answer is no, since 1 is a valid solution of the inverse but is not in the domain of the original function. Hence I would say that f-1 is not a true inverse and thus f is not onto.

If I'm wrong, would you mind explaining why you believe I am wrong.

Thanks,

Timkin

[edited by - Timkin on January 29, 2004 11:55:22 PM]

quote: Original Post by Anonymous Poster
I fully agree with that, pacman2003.

me too.

quote:Original post by Timkin
I think you''ve misunderstood what I wrote. I''m not talking about an inverse function thats not an inverse. I''m talking about situations where x!=f^-1(f(x)). For a function to be onto, it must be the case that x=f^-1(f(x)). If this does not hold true, then the function cannot be onto.

My definition of an inverse function is the mathematical definition we all use. That is, take a function y=f(x). The inverse function of f, denoted f^-1, is obtained by swapping y for x and x for y (giving x=f(y)) and then rearranging to make y the subject again. The final function is f^-1. I''m sure we''d all agree on that, since its a standard definition.

I''m not so sure we all agree on that. I would say that according to standard definitions, an inverse function f^-1(x) is a function such that x = f^-1(f(x)) for all x in the domain of f(x), ie it directly contradicts the possibility that x != f^-1(f(x)). See for instance Planet Math: http://planetmath.org/encyclopedia/SomethingRelatedToInjectiveFunction.html , or Mathworld: http://mathworld.wolfram.com/InverseFunction.html .

quote:

quote:Rather, the typical way to put it seems to be that a function is not ''onto'' if the range differs from the codomain.

...and how would you describe that mathematically?

I would say that there is at least one y in the codomain for which there is no x such that y=f(x). Whether f(x) has an inverse or not has nothing to do with it.

quote:
First, let''s make it clear that there is obviously a problem with the function as presented. Clearly it does not map R ->R . In fact it maps R /{-0.25,1}->R /{-0.25}.

Actually, it maps R /{-0.25,1} -> R /{-0.25,-1}.

quote:
I may have made an error, however assuming I haven''t, then the inverse function I get is:

f<sup>-1</sup>(x) = 3(x-1)+-5(x+1)         -------------            2(4x+1)    

The domain of this function excludes the point -0.25, so the domain of the inverse coincides with the range of the function, namely R /{-0.25}. According to your definition, that makes the function onto. However, is it the case that x=f^-1(f(x))? In other words, does the range of the inverse coincide with the domain of the function? The answer is no, since 1 is a valid solution of the inverse but is not in the domain of the original function.

This is such a mess I don''t know where to begin... However, I don''t know if any of it really matters, after reading the following:

quote: Hence I would say that f^-1 is not a true inverse and thus f is not onto.

So now you say that f^-1 is not really an inverse after all?? And then you say that THUS f is not ''onto''! What is that supposed to mean? That only functions that are invertable can be ''onto''? That only bijective functions can be surjective??