Re: onto function viewpoint
- From: Narcoleptic Insomniac <i_have_narcoleptic_insomnia@xxxxxxxxx>
- Date: Sun, 25 Dec 2005 00:58:56 EST
On Dec 24, 2005 2:05 AM CT, Gary Weselle wrote:
> Hello
Hi.
> Do I understand an onto function correctly?
I dunno.
> The book I am studding math from showing a figure,
> which indicates that 2 points from the domain going to
> the same point in the co-domain. Which makes me think
> that an onto f(X) = Y function is this:
>
> For all Y there exists n Xs "that is n number of x"
> such that n >= 1.
For all Y? There is only one Y; I think you meant to say
something like:
For all y in Y there exists n x's in X "that is n number
of x" such that n >= 1.
The upper case letters usually denote sets & spaces,
whereas the lower case letters usually denote elements
or points in those sets/spaces.
Your idea is basically right; every element in the range
has one (or more) pre-image(s) in the domain.
> But in the definitions in the next page. It says
> something not quite like that, it says;
>
> A function f: X -> Y is onto (or surjective) if for
> every y in Y there exists x in X such that f(x) = y.
Yes, this is the usual definition for a surjective
function.
> It seams to me, "if I understand correctly" that it
> should say (such that f(one or more x) = y.
Well, the above definition allows for the possibility
of one or more pre-images of y. By that I mean the
definition doesn't say anything about x being unique or
the only x such that f(x) = y.
> Which is correct?
The usual definition is correct, but you seem to have the
right idea of what is going on. Here is an example:
Let X = {Gary, Kyle, William, Robert, Ted} and let
Y = {Sarah, Nicole, Melissa}. Now define the function
f: X --> Y such that
f(Gary) = Sarah
f(Kyle) = Nicole
f(William) = Mellisa
f(Robert) = Nicole
f(Ted) = Sarah.
This function f is surjective because for any girl you
pick in Y there is a guy in X such that f(guy) = girl.
Specifically,
f^(-1)(Sarah) = {Gary, Ted}
f^(-1)(Nicole) = {Kyle, Robert}
f^(-1)(Mellisa) = {William}.
Note that although this function f is surjective, it's
not injective because f(Gary) = f(Ted), but Gary =! Ted.
> Thanks
I hope that helps a bit.
Regards,
Kyle
.
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