# Are the discrete and indiscrete functors the only sections of the forgetful functor from topological spaces to sets?

A section of the forgetful functor from topological spaces to sets assigns a particular topology $$\tau_X$$ on each set $$X$$ in such a way that every function $$X \to Y$$ becomes a continuous function $$(X,\tau_X) \to (Y,\tau_Y)$$.

Two such sections are the discrete topology (with $$\tau_X=P(X)$$, the full power set of $$X$$) and the indiscrete topology (with $$\tau_X=\{\emptyset,X\}$$). This leads to the following question:

Are those two sections the only sections of the forgetful functor from topological spaces to sets?

There are exactly four topologies on a two-element set. Besides the discrete and indiscrete topologies, the two other topologies on $$\{0,1\}$$ are $$\{\emptyset,\{0\},\{0,1\}\}$$ and $$\{\emptyset,\{1\},\{0,1\}\}$$.

Each topology on $$\{0,1\}$$ classifies a different kind of subset:

• A map from $$X$$ to $$\{0,1\}$$ with the discrete topology is continuous if and only if it is the characteristic function of a clopen subset of $$X$$.
• A map from $$X$$ to $$\{0,1\}$$ with the indiscrete topology is always continuous.
• A map from $$X$$ to $$\{0,1\}$$ with the topology $$\{\emptyset,\{0\},\{0,1\}\}$$ is continuous if and only if it is the characteristic function of a closed subset of $$X$$.
• A map from $$X$$ to $$\{0,1\}$$ with the topology $$\{\emptyset,\{1\},\{0,1\}\}$$ is continuous if and only if it is the characteristic function of an open subset of $$X$$.

It follows that if $$\tau_{\{0,1\}}$$ is not the indiscrete topology, then $$\tau_X$$ must be the discrete topology for all sets $$X$$. So, besides the discrete topology functor, all sections of the forgetful functor from topological spaces to sets must assign the indiscrete topology to a two-element set. Also, clearly, there is exactly one topology on any empty or singleton set, and it is both discrete and indiscrete.

• Nice observation! An alternative analysis is that topologies on $\{0,1\}$ are collections of endomorphisms of $\{0,1\}$, that a topology induced by a section has to be a right-ideal of the monoid of all endomorphisms of $\{0,1\}$, that such ideals are the ones consisting of all functions (discrete topology), of the constant functions (indiscrete topology), and of no functions (but the last is not a topology). Commented Jan 31, 2022 at 4:14

I think your idea with the space $$\{0, 1\}$$ can be generalized. Imagine that there is a space $$Y$$ such that the associated topology is not indiscrete, i.e. there is a nontrivial open set $$V \subset Y$$. Fix some points $$y_0 \in V, y_1 \notin V$$.
Fix a set $$X$$. If $$U \subseteq X$$, consider the function $$f \colon X \to Y, \qquad x \mapsto \begin{cases} y_0 & x \in U \\ y_1 & x \notin U. \end{cases}$$ By assumption, the induced function $$f\colon (X, \mathcal{T}_X) \to (Y, \mathcal{T}_Y)$$ is continuous. Therefore, $$f^{-1}(V)=U$$ is open in $$X$$, and so $$\mathcal{T}_X = \mathcal{P}(X)$$. This shows that the assignment is the discrete topology.
• Nice! More generally, a section of the forgetful functor from the category of objects structured with collections of morphisms to a fixed object $M$ is the functor associating discrete structures if and only if some structure morphism $Y\to M$ of the section functor is a split epimorphism. Commented Jan 31, 2022 at 4:11