This questions arised to me when I realized that $\neq$ is not an equivalence relation ($1\neq 1$ is false, and $1\neq 2\neq 1$ is false) and applying the operations leads to contradictions, like $1\neq 2 \implies 1\cdot 0\neq 2\cdot 0$ is false. So, what part of the definition of the functions let us apply it to the equivalence relations without leading us to contradictions? This reminded me that all we do with functions, if it's not well-defined it doesn't work. I got the definition of the function from EQUIVALENCE RELATIONS, WELL-DEFINEDNESS... - ALLAN YASHINSKI pdf, which is the following.
A function $f$ from a set $X$ to a set $Y$, is a subset of $X\times Y$ with the following property: $$ (\forall x\in X)(\exists! y\in Y)[(x,y)\in f] $$ [...]. This means, for each $x\in X$, there exists a unique $y\in Y$ such that $(x,y)\in f$. Here, $f$ is a set and $(x,y)\in f$ is equivalent to $f(x)=y$.
Why we want to exists a unique $y\in Y$ instead that only exists, without uniqueness; i.e., to what call well-defined?
I think this might be related of, what leads us to jump from binary relations to functions? And which part of working with functions and equivalence relations breaks without needing to be well-defined?
What on earth do you mean? You can't apply functions to equivalence relations.
Take the equivalence relation on $\mathbb R$, $a$ is equivalent to $b$ if $a-b$ is an integer. So $1.25 \cong 2.25 \cong 37.25$. And take the function $f(x) = x^2$. Then $f(1.25) = 1.5625$ and $f(2.25) =5.0625$ but $5.0625-1.5625 = 3.5$ is not an integer so $1.5625\not\cong 6.0625$