# Different Hausdorff topologies with the same continuous mappings

Fix a set $$X$$. For a topology $$\tau$$ on $$X$$, let $$F_\tau = \{f: X \to X \mid f \text{ continuous w.r.t. } \tau\}$$. What would be an example of a set $$X$$ and two Hausdorff topologies $$\tau$$ and $$\sigma$$ on $$X$$ such that $$\tau \neq \sigma$$ but $$F_\tau = F_\sigma$$?

• Maybe explore the very limited world of topologies on finite sets, with maybe 4 or 5 elements. Mar 1 at 14:54
• @DanAsimov If the set is finite there is only one Hausdorff topology: the discrete one. So, this doesn't work. Mar 1 at 14:55
• Well, so much for that idea. Here's another possibly bad idea: What if we define an alternative topology on the real line with a base of half-open intervals [a, b) instead of (a, b). Will this have the same continuous maps ℝ → ℝ as the standard topology? Mar 1 at 14:57
• @DanAsimov, I think that one is disconnected, so for example there is a continuous map whose image is exactly $\{0, 1\}$. Mar 1 at 15:07

In more detail: suppose $$(X, \tau)$$ is a nontrivial Hausdorff space with the property that any $$\tau$$-continuous function $$X \to X$$ is either the identity or constant. Let $$f: X \to X$$ be any non-identity bijection of $$X$$ (eg one that swaps two elements). Let $$\sigma$$ be the topology on $$X$$ that makes $$f: (X, \tau) \to (X, \sigma)$$ into a homeomorphism (ie "transport the structure along $$f$$"). Then $$F_\tau$$ and $$F_\sigma$$ both consist only of the identity function and the constant functions. But $$\tau$$ and $$\sigma$$ are not equal, because otherwise $$f$$ would be a continuous function from $$(X, \tau) \to (X, \tau)$$.
• @Soma, that's not so hard - $\sigma$ is just the collection of all sets of the form $f(U)$ where $U \in \tau$. The idea is to use $f$ to "relabel" the points of $X$. See also here. Mar 1 at 15:16