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Homework Help: Proving a Relation Satisfies the Transitive Property

  1. Jun 27, 2014 #1
    1. The problem statement, all variables and given/known data
    suppose [itex]f~:~A \rightarrow B[/itex] be a surjective map of sets. Prove that the relation [itex]a Rb \iff f(a)=f(b)[/itex] is a equivalence relation whose equivalence classes are the fibers of [itex]f[/itex].

    2. Relevant equations

    3. The attempt at a solution
    I was able to easily prove that the relation satisfied the reflexive and symmetric part, but there are a few details of the transitive case that I am uncertain of. Here is some of my work:

    Suppose that [itex](a,b) \in R[/itex] and [itex](b,c) \in R[/itex], meaning that the function [itex]f[/itex] maps the elements a and b to the same element in the codomain, call it [itex]\alpha[/itex], and the function maps b and c to the same element; that is, [itex]f(a) = f(b) [/itex] and f(b) = f(c). Does this immediately imply that [itex]f(a) = f(c)[/itex], or do I have to do some further reasoning, such as reiterating that [itex]f[/itex] is indeed a function, which means that [itex]f[/itex] can only associate one element in its codomain to each element in its domain, suggesting that [itex]f(b) = \alpha = \beta[/itex]?

    For instance, suppose in the problem that [itex]f[/itex] was not specified as a function. Then if [itex]f[/itex] was [itex]f(x) = \pm \sqrt(x)[/itex], then it wouldn't be transitive, right?
  2. jcsd
  3. Jun 27, 2014 #2
    Wait, f is not a function, is it?
  4. Jun 27, 2014 #3
    I also have a few questions relating to the second portion of the question.

    From what I have read, it seems that the range and image are two terms for the same concept. And that the image (range) is to the codomain as the pre-image is to the domain; in other words, just as not every element in the codomain is associated with an element in the domain, not every element in the domain is mapped to an element in the codomain. The pre-image is a subset of the domain. The image and pre-image are the actual elements that f acts upon.

    Do all of these claims appear to be true?
  5. Jun 27, 2014 #4


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    A "map of sets" is a function. In category theory there are other types of maps apparently.
  6. Jun 27, 2014 #5
    Your reasoning for transitivity is correct. You don't need to do anything else.

    I would say that is accurate. Apparently some people use "range" to refer to the codomain, but I've never experienced it. I just avoid using the word.
  7. Jun 27, 2014 #6


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    As far as I know, the pre-image is the domain, there is no difference. I believe it changed before 1960 (I have read quite a few old books), some time before that one could speak of partial functions or total functions. But not any more, a function is always total, defined on the whole domain.
  8. Jun 27, 2014 #7
    Edit: I disagreed at first, but upon looking into this, the convention seems to be all over the place. The majority seems to agree with verty, though.
    Last edited: Jun 27, 2014
  9. Jun 28, 2014 #8


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    Software functions can be partial or total, depending on the preconditions, but mathematicians seem to assume that functions are always total, in my experience. I've paid attention to this because I liked the idea that one could have partial or total functions, but in the books I've read, a mathematical function is always total.

    Even in computer science, one might like to say that a software function is partial but the mathematical function being implemented is total. In no recent book have I seen mathematical functions being called partial.
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