Brouwer's fixed point theorem and the one-point topology

In the 2-dimensional case, Brouwer's fixed point theorem (BFPT) says that every continuous function $D^2\to D^2$ has a fixed point, where $D^2$ is the disk.

Now fix a particular topology: pick some point $x_0\in D^2$ and use it to define the one-point topology $\cal T_0$ on $D^2$: it includes all sets $A$ with $x_0\in A$, and the empty set. (This is indeed a topology, see for example https://en.wikipedia.org/wiki/Particular_point_topology).

With respect to $\cal T_0$, a self map $D^2\to D^2$ is continuous if and only if it is constant or has $x_0$ as a fixed point. So, for every self map on $D^2$, continuity with respect to $\cal T_0$ means that a fixed point exists. Hence the BFPT is trivially true, by definition of $\cal T_0$.

In conclusion, there are topologies where BFPT is a theorem that requires proof, and there is a topology $\cal T_0$ where BFPT is true simply by definition.

This gives $\cal T_0$ a special place among all possible topologies on $D^2$: it is the topology that makes BFPT trivial. Does such a situation or property have a name? Does it have a category theory interpretation (maybe like "universal property")?

I feel there is a certain equivalence between BFPT and $\cal T_0$ here. They characterize each other in a certain way: $\cal T_0$ makes BFPT trivially true by definition, and BFPT links continuity and fixed points (like $\cal T_0$ does). Can this sense of equivalence be expressed rigorously?

EDIT: I want to thank you for the comments which were very helpful for me. I still seem to struggle with the question in my head, so I edited my question with the intention to make it more precise.

Solution 1:

I think there is a way to make this rigorous. To do so, you should keep in mind that your topology $\mathcal T_0$ is not well-defined. Instead it depends on a parameter, namely a choice of a point $p \in D^2$ (what you denoted $x_0$).

To indicate this, let me append $p$ to the definition: $$\mathcal T_0(p) = \{A \subset X \mid p \in A \quad\text{or}\quad A = \emptyset\} $$

So with this in mind, perhaps there is an equivalence going on: the Brouwer fixed point theorem is equivalent to the statement that for each function $f : D^2 \to D^2$, if $f$ is continuous in the standard topology then there exists $p \in D^2$ such that $f$ is continuous in the topology $\mathcal T_0(p)$.

Now, I would not say there is anything deep going on here, the topology $\mathcal T_0(p)$ has been concocted to make this equivalence almost trivial.