When are the quotient topology and the final topology not the discrete topology?

Question

It says here that the quotient topology is the final topology on the quotient space with respect to the map $q$.

The link seems to say that the final topology is the finest topology that makes those functions continuous.

It seems to say here that the finest topology is always the discrete topology.

I'm not adding something up correctly here because if I were, Wikipedia would simply say the quotient topology is the discrete topology, which is also the final topology and two names out of three would be redundant.. What am I missing?

Answer 1

Consider $A=\{a,b,c\}$ and the map $p:\mathbb R\to A$ defined by $$ p(x) = \begin{cases} a,& x>0\\ b,& x<0\\ c,& x=0. \end{cases} $$ Let $\mathcal T$ be the quotient topology. Recall that $\mathcal T = \{U\subset A: p^{-1}(U) \text{ open in } \mathbb R$}. We see that \begin{align} p^{-1}(\{a\})&=(0,\infty),\\ p^{-1}(\{b\}&=(-\infty,0),\\ p^{-1}(\{a,b\}) &= \mathbb R\setminus\{0\},\\ p^{-1}(A) &= \mathbb R \end{align} are all open, and $p^{-1}(S)$ is not open for any other subset of $A$. So $$\mathcal T = \{\{a\},\{b\},\{a,b\},A\}\ne 2^A, $$ that is, the quotient topology is not the discrete topology on $A$.

Answer 2

You must not confuse domain and range.

In your second link you have a function $f : S \to S$ from a set $S$ to itself. Then the discrete topology is the finest topology on $S$ such that $f$ becomes continuous.

The is just a special case of the following:

If you have a function $f : S \to Y$ from a set $S$ to a topological space $Y$, then the discrete topology is the finest topology on the domain $S$ such that $f$ becomes continuous.

For the quotient topology you have a (surjective) map $f : X \to S$ from a topological space $X$ to a set $S$ and look for the finest topology on the range $S$ such that $f$ becomes continuous. This is the final topology on $S$ with respect to $f$. In general this will not be the discrete topology.

A simple mnemonic aid could be this: You have a collection of functions $f_\iota : X_\iota \to S$ from topological spaces $X_\iota$ to a set $S$. Thus $S$ is the "end" (in Latin "finis") of all arrows, and the final topology for this "finis" should contain as many open sets as possible to make all $f_\iota$ continuous. This assures that all $X_\iota$ contribute as much as possible to the topology on $S$.

Answer 3

The function $q: X \rightarrow Y$ is pre-given, and the demand on the topology on the codomain is that the inverse images of open sets under $q$ are open. So it is quite rare for the discrete topology on $Y$ to obey that: it means precisely that all fibres $q^{-1}[\{x\}]$ (the equivalence classes if $q$ is defined from an equivalence classes) are open in $X$ (as singletons form a base for the discrete topology on $Y$). If that happens to be the case, the discrete topology is the finest topology that makes $q$ continuous. But more often than not, the inverse images of sets under $q$ are not all open (it will depend on the topology on $X$ which is also pre-given); the final topology (or quotient topology, which is the usual name if it is about just one (surjective) function) is just $\{A\subseteq Y\mid q^{-1}[A] \in \mathcal{T}_X\}$, and will generally be smaller (coarser) than the discrete topology.