What is " the equivalence relation associated to $f$" , $f$ being a surjective mapping from a group $G$ and a set $E$.

Question

Context : Equivalence relations and quotient groups.

Source : Reversat & Bigonnet, Algèbre pour la licence ( Undergraduate abstract algebra), 1997, p. $31$ .

The notion refered to in the title appears in the statement of the following theorem :

Let $(G, \bullet)$ be a group, $E$ a set , and $f$ a surjective function from $G$ to $E$.

If the equivalence relation associated to $f$ is compatible with $(G, \bullet)$'s structure , then there is a group structure on $E$, denoted by $(E, \star)$, such that $f$ becomes a homomorphism from $(G, \bullet)$ to $(E, \star)$.

The proof of this theorem begins like this : " We are going to prove that the relation $f(x)f(y)=f(xy)$ defines a composition law [ an operation] on $E$... "

My question :

is the relation refered to in the proof, and defined by $$x\mathscr R y \iff f(x)f(y)=f(xy), $$

the same relation as the equivalence relation refered to in the theorem itself?

If it is the case, $\mathscr R$ has to qualify as an equivalence relation ( reflexivity, symmetry, transitivity).

How can this be shown?

And before showing this,

how to understand the expression " equivalence relation associated to function $f$"?. Is it an equivalence relation on $G$ or on $E$? Does $f$ admit of only one possible equivalence relation associated to it?

The source quoted above contains a " Prerequisites" chapter, but the notion of " equivalence relation associated to a function " is not defined therein.

Answer 1

No, the equivalence relation associated to $f$ is the relation $\sim$ on $G$ defined by $x\sim y$ iff $f(x)=f(y)$. This is just what "the equivalence relation associated to a function" is defined to mean (at least in this context). That phrase does not have any meaning absent a specific definition, so it does not make sense to ask whether $f$ admits "only one possible equivalence relation associated to it".

The relation referred to in the statement is also different from what you think it is. It does not make sense to define a relation $xRy$ by $f(x)f(y)=f(xy)$ because you have not yet defined what $f(x)f(y)$ even means. (Remember, $E$ is just a set and $f(x)$ and $f(y)$ are elements of it, and we do not yet have a binary operation on $E$ that would give a meaning to $f(x)f(y)$.) Instead, the relation referred to is a relation $R$ from $E\times E$ to $E$ defined by $(a,b)Rc$ iff there exist $x,y\in G$ such that $a=f(x)$, $b=f(y)$, and $c=f(xy)$. The point here is that if we knew this relation $R$ were actually a function, then we could think of it as a binary operation $\star$ on $E$ that would satisfy $f(x)\star f(y)=f(xy)$. In this sense this $R$ is "the relation $f(x)f(y)=f(xy)$" (which admittedly is a rather confusing phrase to use to define this $R$).

Answer 2

Associated to every function $f : A \to B$ (surjective or not) there is an equivalence relation on its domain $A$ defined by $$a_1 \sim a_2 \iff f(a_1)=f(a_2) $$ The equivalence classes of this relation are precisely the inverse images of the individual points of the subset $\text{image}(F) \subset B$, i.e. for each $b \in \text{image}(F)$ the nonempty set $f^{-1}(b) = \{a \in A \mid f(a)=b\}$ is an equivalence class.

I don't know and can't guess what role the relation $\mathscr R$ has to play in what you are reading, but it is definitely not the same as the equivalence relation associated to $f$.