Set of equivalence classes as the image of a map

Question

In the book I am reading it states that any map of sets $f\colon S \to T$ gives us an equivalence relation $\bar{S}$ defined by $a\sim b$ if and only if $f(a)=f(b)$. This all makes sense to me, but then the book states how the set $\bar{S}$ of equivalence classes is equal to the image of the map. Each equivalence class is a subset of $S \times S$, so how exactly is this the case? Does the author mean that each equivalence class could basically be represented by $f(a)$, that is, represent $[a]$ as $f(a)$ as every $x \in [a]$ maps to $f(a)$ by definition?

Any help is appreciated.

The set of equivalence classes is not "equal" to the image, but there is a natural bijection with the image, by mapping the equivalence class of $a$ to the element $f(a)$ of the image. — Arturo Magidin, Apr 08 '24 at 01:52
Note that while an equivalence relation on $S$ is certainly a subset of $S\times S$, an equivalence class is not—it's just a subset of $S$ itself (indeed, the equivalence classes form a partition of $S$). — Greg Martin, Apr 08 '24 at 03:13
Such relations are called (equivalence) kernels. We call $, \sim,$ the kernel of $\rm,f.,$ The equivalence classes $,f_c = f^{-1}(c),$ are called the fibers or preimages, or level sets / curves of $f. \ \ $ — Bill Dubuque, May 15 '24 at 15:41

score 2 · Answer 1 · answered May 15 '24 at 10:43

The implication goes both ways i.e., the following statements are equivalent:

$R$ is an equivalence relation on $A$
There is a set $B$ and a function $f:A\to B$ such that $xRy\Leftrightarrow f(x) = f(y)$.

The canonical way to prove the implication $1.\Rightarrow 2.$ is, I suppose, to go with the function: $$[\_]_R:A\to\mathcal{P}(A)\\ a\mapsto[a]_R$$

Maybe the book's author tried to express that the function $[\_]_R$ is one (the canonical) way to obtain such a map.

Set of equivalence classes as the image of a map

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