Let $f:A\rightarrow B$ be a surjective map of two non-empty sets $A$ and $B$. We define $a\sim b$ if $f(a)=f(b)$. Prove that this is an equivalence relation with the fibers of $f$ as its equivalence classes.

Here is my attempt at the proof:

Since $f$ is a function, $\forall a\in A$, $f(a)=f(a)$, and therefore we have reflexivity. Given $a,b\in{A}$, if $a\sim b$, $f(a)=f(b)$, meaning $b\sim a$, thus we have symmetry. Given $a,b,c\in{A}$, if $a\sim b$ and $b\sim c$, $f(a)=f(b)$, and $f(b)=f(c)$, so $f(a)=f(c)$ then $a\sim c$. Therefore, we have transitivity. The fiber of $f$ is defined as $f^{-1}(b)=\lbrace{a\in A|\space{f(a)=b}}\rbrace$, for some $b\in B$. An equivalence class is defined as a subset of the form $\lbrace{a_1\in A|\space a_1\sim a_2}\rbrace$, for some $a_1, a_2\in A$. The fiber of $f$ contains all $a_i\in A$ such that $f(a_i)=b$, thus are equivalence classes of similarity.

I want to know if there are

$1)$ Any corrections to be made, and

$2)$ Better wording for the proof.

Thank you all in advance.


The first part seems fine. I would write somewhere how you are really using the fact that "$=$" is an equivalence relation to obtain each of the steps. Also it would be better to explicitly show that, for instance, you are making the jump from $f(a)=f(b)$ and $f(b)=f(c)$ to $f(a)=f(c)$ before concluding $a\sim c$.

For the second part, it's not clear what you are saying. Write down what you intend to prove. Also, did you use surjectivity? If not, then why is that a hypothesis?

You are asked to show that the fibers of $f$ are the equivalence classes of this relation. This means we need to prove that each fiber of $f$ is an equivalence class, and each equivalence class is a fiber.

Let $b\in B$ and consider the fiber $f^{-1}(b)$. We want to show that this is the equivalence class of some element of $A$. But which element? Details in spoiler:

Since $f$ is onto, there exists $a\in A$ such that $f(a)=b$. Then the equivalence class $[a]$ of $a$ equals $f^{-1}(b)$. Indeed, we have $c\in [a]$ iff $c\sim a$ iff $f(c)=f(a)$ iff $f(c)=b$ iff $c\in f^{-1}(b)$.

Conversely, let $a\in A$ and consider the equivalence class $[a]$ of $a$. We want to show that this is the fiber of some element in $B$ under $f$. But which element? Details in spoiler:

We claim that $[a]$ equals the fiber $f^{-1}(f(a))$ of $f(a)$ under $f$. This follows because $c\in [a]$ iff $c\sim a$ iff $f(c)=f(a)$ iff $c\in f^{-1}(f(a))$.

  • $\begingroup$ Thank you for the detailed response—I'm not too sure how I can further explain the jump from $f(a)=f(b), f(b)=f(c)\rightarrow f(a)=f(c)$. Can I just use my original sentence if I state that "$=$" is an equivalence relation? $\endgroup$ – 高田航 Sep 30 '17 at 0:52
  • $\begingroup$ @高田航 I'm sorry! Having reread your proof I noticed that you did in fact write $f(a)=f(c)$ . It's when proving symmetry that you make the jump from $f(a)=f(b)$ to $b\sim a$ without first noting $f(a)=f(b)$. It's not necessary to write every little step, but it does make the proof easier to follow. Later on in your mathematical career you can get away with saying something like "reflexivity of $\sim$ follows by reflexivity of $=$", and likewise for the other properties, but for now it's better to be careful and explicitly show every step. $\endgroup$ – John Griffin Sep 30 '17 at 0:56
  • $\begingroup$ Ah I see, thank you very much. One last thing: do I have to prove the properties of equivalence relations for "$=$", or can the definition of "$=$" be assumed? $\endgroup$ – 高田航 Sep 30 '17 at 0:59
  • $\begingroup$ @高田航 The axioms defining "$=$" make it an equivalence relation, so it's not something you will (or can) prove. $\endgroup$ – John Griffin Sep 30 '17 at 1:03
  • $\begingroup$ Ok, thank you again for the detailed answer. $\endgroup$ – 高田航 Sep 30 '17 at 1:03

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