Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\omega_X$ be the set of all topologies on $X$. Given $f:X\rightarrow X$, define $R_f \subset \omega_X \times \omega_X $ as those pairs of topologies on $X$ which make $f$ continuous. For example $\left(\text{Discrete Topology},-\right)$ or $\left(-,\text{Indiscrete Topology}\right)$ are always in $R_f$. But when $f$ can be uniquely determined, by its $R_f$? Here is one such case: $$ \forall x \in X: f(x)=x \iff R_f= \left\{ \left(T_\alpha,T_\beta\right)\subset \omega_X \times \omega_X | T_\beta \subset T_\alpha \right\}$$

Can some one give me more elaborative examples of this please?

share|cite|improve this question
up vote 7 down vote accepted

$R_f$ uniquely determines $f$ for any non-constant function $f:X \to Y$ (if, and only if, $f$ is constant, $R_f$ is $\omega_X\times\omega_Y$).

Proof: Fix $y\in Y$. Define the topology $T_y=\{\emptyset,Y,\{y\}\}$. Then $(T,T_y)\in R_f$ iff:

  • $f^{-1}(\emptyset)=\emptyset\in T$ (always true)
  • $f^{-1}(Y)=X\in T$ (always true)
  • $f^{-1}(\{y\})\in T$

Take the intersection of all such $T$: it is the set $T_0=\{\emptyset,X,f^{-1}(\{y\})\}$. Because $f$ is not constant, $f^{-1}(\{y\})\ne X$ and so $f^{-1}(\{y\})$ is the largest element of $T_0\setminus\{X\}$.

Since $f$ is uniquely determined when $f^{-1}(\{y\})$ is given for all $y$, this proves the result.

(Note: we only needed to use topologies with a finite number of open sets.)

share|cite|improve this answer
I think that you want $X$, not $Y$, throughout the paragraph containing the definition of $T_0$. – Brian M. Scott Jun 17 '12 at 10:58
Absolutely, thanks. – Generic Human Jun 17 '12 at 11:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.