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I posted this question last week on Math SE and got upvotes and helpful comments that allowed me to make the question more precise https://math.stackexchange.com/q/3765546/810513. As I did not get an answer to this question, please allow me to ask it on MO:

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 just 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?

I am struggling to express this property and my sense of "equivalence" more precisely and thus understand it better - I would be very grateful if someone could help me with this.

EDIT: Thank you for great comments! They are extremely helpful. I edited and corrected my question to reflect them.

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    $\begingroup$ "Continuity and having the fixed point $x_0$ are one and the same thing here.", that's not true, is it? For example if $x_1$ is any other point, the constant map to $x_1$ is also continuous with respect to your $x_0$-topology, since all preimages are either empty or everything. But this does not have $x_0$ as fixed point. $\endgroup$
    – Achim Krause
    Jul 26, 2020 at 15:16
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    $\begingroup$ Maybe the problem is that literally as written, your topology is not a topology, since the empty set also needs to be open. $\endgroup$
    – Achim Krause
    Jul 26, 2020 at 15:18
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    $\begingroup$ It is still not true that continuity and having fixed point $x_0$ are one and the same thing for that topology. For any topology any point is the unique fixed point for some continuous map (e. g. the constant map with value this point). $\endgroup$
    – მამუკა ჯიბლაძე
    Jul 26, 2020 at 17:33
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    $\begingroup$ I don't sense any real equivalence here between continuity and having a fixed point, since there are many non-constant self maps which fix other points but don't fix $x_0$. I think the best we can say is "BFPT is easier to prove when our topology is a lot simpler." My guess is that there isn't anything deeper going on. $\endgroup$
    – Matt Feller
    Jul 26, 2020 at 17:54
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    $\begingroup$ The first paragraph is perplexing to me. You write "This is valid for any choice of topology." But the BFPT is a fact about the standard Euclidean topology on $D^2$, and certainly fails for other topologies. To be explicit, consider the indiscrete topology on $D^2$, in which only the empty set and $D^2$ itself are open. Every function $D^2\rightarrow D^2$ is continuous in this topology, but there are certainly functions from $D^2$ to itself with no fixed point (e.g. a function that maps the origin to $(0,1)$ and rotates the rest of the disk by 90 degrees). $\endgroup$
    – Dan Ramras
    Jul 27, 2020 at 5:43

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