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

Question

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.

Answer 1

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.

We get that the discrete and indiscrete functors are the only sections of the forgetful functor, as claimed.