I'm trying to understand a point in my professor's lecture notes on equivalence relations.
First, we define a function $f: S \to T$ with $a \sim b$ if and only if $f(a) = f(b)$. This is surely an equivalence relation, and the partition of $S$ is given by $S = \bigsqcup\limits_{t \in \mathrm{Im}(f) \subset T} f^{-1} (t)$. We can then say that $f$ factors through the quotient via the composition of maps \begin{align*} \underset{a}{S} \underset{\longmapsto}{\twoheadrightarrow} \underset{[a]}{S/\sim} \; \; \underset{\longmapsto}{\hookrightarrow} \underset{f(a)}{T}. \end{align*} so if $\pi$ is the canonical projection map and $\varphi$ the map $[a] \to f(a)$, then $f = \varphi \circ \pi$.
The next point he makes doesn't make any sense to me. He writes: \begin{align*} \text{equivalence relation on $S$} \iff & \text{ partition of $S$ into disjoint subsets } \\ \iff & \text{ surjective map from $S$ to another set $T$} \\ & \text{(up to composition with a bijection $T \stackrel{\sim}{\to} T'$)}. \end{align*} The first biconditional makes sense. If we have an equivalence relation on $S$, we get a partition of $S$ into equivalence classes. Conversely, given a partition of $S $, we can define an equivalence relation by the rule that $a$ is related to $b$ if they live in the same set in the partition.
I do not understand the argument behind second biconditional or its implication. In the above example, for instance, there is no surjective map from $S \to T$, but from $S \to S/\sim$, the set of equivalence classes, which is then embedded in $T$. I'm also not sure what he means to "up to composition with a bijection $T \to T'$, since $T$ is by definition the codomain of $f$.
UPDATE: After thinking about this more and reading the very helpful answers below, I think what he is saying is that if there is some other function $G: S \to T$ that induces the same equivalence relation, we can "compose" with the bijection $T \to T'$ sending $f(a)$ to $g(a)$. The equivalence classes are determined by the fibers, so we don't require $f(a) = g(a)$ for every $a$. Is this the correct idea?
The composite map from $S$ to $S / \sim$ to $T$ (which in fact equals $f$) is surjective.
Your professor means the following:
Given a particular equivalence relation $\sim \subseteq S^2$, there are potentially many choices $f : S \to T$ of surjective functions such that $\{(a, b) | f(a) = f(b)\} = \sim$.
For example, consider the set $A = \{1, 2\}$ and the equivalence relation $\{(1, 1), (2, 2)\}$. This equivalence relation is induced by
$x \mapsto x : A \to A$
$x \mapsto x + 1 : A \to \{2, 3\}$
$x \mapsto 2 - x : A \to A$
etc.
You'll notice that all of these functions have something in common: their codomain is in bijection with $A$.
We consider two functions $f : A \to B$ and $g : A \to C$ to be equivalent if there is a bijection $h : B \to C$ such that $h \circ f = g$. Note that this is in fact an equivalence relation. Furthermore, note that if $f$ is surjective and $g$ is equivalent to $f$, $g$ is also surjective. Finally, note that surjective functions can be equivalent "in at most one way" (meaning that there is at most one bijection $h : B \to C$ such that $h \circ f = g$).
Your professor is stating that if two surjective maps $f : A \to B$ and $g : A \to C$ give the same induced equivalence relation $\sim$, then the two maps are equivalent under the above definition. Conversely, if the two maps are equivalent under the above equivalence relation, they will produce the same equivalence relation on $A$.
The "canonical" way of taking an equivalence relation $\sim$ and getting a surjective function $f$ is by taking the projection map $f : A \to A / \sim$, which is defined using equivalence classes.