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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?

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1 Answer 1

up vote 6 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.)

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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

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