Finest topology such that a map is continuous : isn’t it the discrete topology

Question

Let $X$ a set and $\mathcal R$ a relation of equivalence of $X$. Set $$q: X\to X/\mathcal R,$$
the natural projection. The quotient topology, it the finest topology such that $q$ is continuous. Isn't it the discrete topology ? Why do we take the finest instead the thickest (as we usually do) ? For me a set $U\in X/\mathcal R$ is open if $q^{-1}(U)$ is open (and I would say that it's the thickest topology s.t. $q^{-1}(U)$ open…). So why do me defined it as the finest ?

Answer 1

It happens that, in general, if you endow $X/\mathcal R$ with the discrete topology, then $q$ will not be continuous. For instance, if $x\mathrel{\mathcal R}y\iff x=y$, then $q$ will be continuous with respect to the discrete topology on $X/\mathcal R$ if and only if $X$ is a discrete topological space.

Note that if $f$ is a map from $X$ into a topological space $Y$ such that $x\mathrel{\mathcal R}y\implies f(x)=f(y)$ then $f$ induces a map $f^\star\colon X/\mathcal R\longrightarrow Y$. With the topology on $X/\mathcal R$ defined as the finest topology such that $q$ is continuous, then $f^\star$ is continuous if and only if $f$ is continuous.