If I say [math]f:A \rightarrow B[/math] , f maps from A to B, is A the codomain or the preimage?
Give me a straight answer please. I want to settle this.
Given that [math]f:A \rightarrow B[/math] , are there elements [math] a [/math] of [math] A [/math] for which [math] f(a) [/math] does not exist in [math] B [/math] or aren't there?
>>8057602
*last question should ask "can there be" not "are there."
>>8057614
So the codomain of a relation equals the pre-image?
>>8057602
Both, but you prob mean codomain
And no there can't.
Say you had some subset [math] C \subset B [/math]
Then the pre-image of f restricted to C is the subset of A [math] \{ x\in A : f(x) \in C \} [/math]
That's how I understand it
>>8057618
Shit I messed up, I meant to ask "is A the domain or the pre-image?"
Co-domain would be B. I should just delete this thread and give up.
So if the pre-image of f restricted to C is the subset of A, then isn't A itself called the "codomain" and this subset, like you said, is the pre-image?
From your example, C is the "image" right? And B is the co-domain. A is the domain, and the subset of A you defined is the preimage. A, B, C, and the preimage are all not necessarily equal.
But if f maps elements of the preimage to elements of C, where does it map elements of A which are not in the preimage, or does it not map them at all? Or, if A and the preimage must be one and the same, then the preimage always equals the domain? That's the question I meant to ask. My bad.
>>8057602
codomain is the complete set in which the domain is mapped to. Not every point in the codomain necessarily has an element that maps to it. The image set or range is the subset of codomain (the elements in set B) that have elements in set A that map to it.
There can be elements in A that have no image in B. Go back to basic functions which operate on the Real number line set.
f(x) = 1/x has no image for the element x=0. x=0 in this case is in the domain, but not in the pre-image. The pre-image is the subset of A that maps to an element in the co-domain.
So in summary:
Domain is the set of possible elements that may be mapped to B. The pre-image is the subset of B that actually have elements that are mapped to B.
The Co-Domain is the set being mapped to. The range is the subset of the co-domain that actually have element in the pre-image that map to them.
>>8057638
So A is the domain, not the preimage, and in general a domain does not have to equal the preimage. Got it. Thanks.
>>8057632
>Shit I messed up, I meant to ask "is A the domain or the pre-image?"
Salright. I messed up a bit to.
f is not restricted to C . f^-1 is...sorta
A is the codomain of f and the pre-image of B under f
My set (calling it S now) is the preimage of C under f.
C isn't really the image. It could be the image of f restricted to S but alcohol is inhibiting my brain.
The image of f (restricted to A) would be a set [math] f[A] = \{ y\in B : x\in A \land f(x)=y \} [/math]
Also remember there is disagreement in some of these definitions.
> But if f maps elements of the preimage to elements of C, where does it map elements of A which are not in the preimage, or does it not map them at all?
The image of S should be in C. The mapping of x not in S, but are in A, f(x) is in B but not C.
> Or, if A and the preimage must be one and the same, then the preimage always equals the domain?
Def not.
>>8057665
So going to the example given by someone above, if f(x) = 1/x , is it fair to say that [math] f: R \rightarrow R [/math], from the reals to the reals?
>>8057688
The domain and codomain are defined as part of the function. So,
[math]f:\mathbb{R}\to\mathbb{R} ; f(x)=\frac{1}{x}[/math] is a different function than [math]f:\mathbb{Z}\to\mathbb{R} ; f(x)=\frac{1}{x}[/math]
>>8057688
if f(x) = 1/x, then f can't map R to R. We could have f:R\{0}->R, however.
>>8057710
And in the first one, R is the domain, but R - {0} is the pre-image, I think. Same for Z in the second.
So the pre-image of a function could just be defined as the smallest possible domain of the function, and presumably the image/range could be defined as the smallest possible codomain of the function.
>>8057717
>R is the domain, but R - {0} is the pre-image
The phrase "the pre-image" without qualification is meaningless. You would say "the preimage of C" where C is a subset of the codomain. It means everything in the domain which maps to C. Also, R is not the domain of the function, because the function is not defined at 0. If we were to pick a value for 0, then the domain would be R. The domain is simply the set of all "inputs" to your function. You can't miss values in the domain like you can in the codomain, i.e. there's no corresponding notion of a range.
>>8057602
>is A the codomain or the preimage?
A is the codomain.
If [math] f(A) = C \subset B [/math], then you'll say [math] A [/math] is the preimage of [math] C [/math].
>>8057725
Domain
>>8057665
>A is the codomain of f and the pre-image of B under f
Wouldn't A be the codomain of f^-1, and the domain of f?
>>8057730
f^{-1} is not usually itself a function. The notation [math]f^{-1}(C)[/math] for pre-image is admittedly confusing.
>>8057730
f^-1 doesn't have to exist if there isn't a bijection. that much I do know.
>>8057725
wouldn't it be correct to say "A is the preimage of of f^-1 under C"?
>>8057764
No?
A is the pre-image of f under B
S is the pre-image of f under C and
If f is bijective.... C is the pre-image of f^-1 under S?
>>8057724
>You can't miss values in the domain like you can in the codomain, i.e. there's no corresponding notion of a range.
That answers my question actually, assuming it's right. yeah.
