For any topological space $(X,\tau)$ we define $$R_{im}(X,\tau) := \{(x,y)\in X^2: (\exists f:X\to X) \text{ continuous and surjective with } f(x) = y\}.$$ Clearly, $R_{im}(X,\tau)$ is reflexive. This relation is also transitive because the composition of two continuous surjective maps is continuous and surjective.

Given a non-empty set $X$ and a reflexive and transitive relation $R$, is there a topology $\tau_R$ on $X$ such that $R_{im}(X,\tau_R) = R$? If not, is the answer positive, if we require $R$ to be

a) an equivalence relation;

b) an ordering relation?

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    $\begingroup$ If $(X,\tau)$ if finite, then $R_{im}(X,\tau)$ must be an equivalence relation. $\endgroup$ – Keith Kearnes Aug 12 '15 at 13:21
  • $\begingroup$ Is it possible to write every equivalence relation $R$ on a finite set $X$ as $R = R_{im}(X,\tau)$ for a suitable topology $\tau$? $\endgroup$ – Dominic van der Zypen Aug 12 '15 at 13:52
  • $\begingroup$ To realize an arbitrary equivalence relation, it suffices to realize the trivial equivalence relation on an arbitrary set (because you can then just replace each point of your space with an indiscrete space to get any equivalence relation with the same number of equivalence classes). $\endgroup$ – Eric Wofsey Aug 12 '15 at 23:36
  • $\begingroup$ @EricWofsey Is that right? The result is true for the two-point space with the identity equivalence relation, but I can't yet see it for an infinite set with two classes, both infinite. Your suggestion to replace each class with the indiscrete space doesn't seem to work in this case. (And using up-sets as in my answer doesn't seem to work either.) $\endgroup$ – Joel David Hamkins Aug 13 '15 at 0:31
  • $\begingroup$ @JoelDavidHamkins: Huh? Why doesn't it work? Concretely, if $A$ and $B$ are disjoint nonempty sets and you topologize $A\cup B$ by saying the only nontrivial open set is $A$, then any continuous surjection from $A\cup B$ to itself must map $A$ to $A$ and $B$ to $B$. $\endgroup$ – Eric Wofsey Aug 13 '15 at 1:13

The answer is no to the general question, and also to question (b), for the following simple reason (which works whether or not the space is finite): if $f:X\to X$ is surjective and $f(x)=y$ for some $x\neq y$, then there must be some $w\neq x$ with $f(w)=x$. Thus, the relation $R_{im}$ must have the property that whenever $x\mathrel{R_{im}} y$ and $x\neq y$, then there must be some $w\neq x$ with $w\mathrel{R_{im}} x$. In particular, $R_{im}$ can have no minimal elements, and so there are numerous counterexamples to the main question and question (b).

Update. Meanwhile, the answer to (a) is positive for finite sets, as asked in the comments:

Theorem. Every equivalence relation on a finite set $X$ arises as $R_{im}(X,\tau)$ for some topology $\tau$ on $X$.

Proof. Suppose that $R$ is an equivalence relation on a finite set $X$. We may place a linear pre-order $\trianglelefteq$ on $X$ in such a way that $x\mathrel{R} y$ is equivalent to $x\trianglelefteq y\trianglelefteq x$. Define the topology $\tau$ to have as open sets exactly the upward closed sets, which have the form $U_x=\{y\in X\mid x\trianglelefteq y\}$, plus the empty set. (This collection is closed under arbitrary unions and intersections.)

I claim that $R_{im}(X,\tau)=R$. First of all, the topology does not distinguish between points within any $R$-equivalence class, and so we may permute within them at will. Thus, $R\subset R_{im}(X,\tau)$. Conversely, suppose that $f:X\to X$ is continuous and surjective. I claim that $f$ is merely permuting within each $R$-class. First, note that $f$ is $\trianglelefteq$-preserving: if $f(x)=y$, then $x$ is in the preimage $f^{-1}(U_y)$, which is open, and so if $x\trianglelefteq x'$, then since $x'$ is in any open set that $x$ is in, it follows that $y\trianglelefteq f(x')$, and so we have $x\trianglelefteq x'\longrightarrow f(x)\trianglelefteq f(x')$. From this, it follows that $f$ must take each $R$-equivalence class into a single $R$-equivalence class. Since $X$ is finite and $f$ is surjective, we know that $f$ is a permutation of $X$, and so $f$ must take the least $R$-class to itself, and the next and so on, since otherwise we'd violate the $\trianglelefteq$-preserving property. In other words, $f$ is merely permuting the points inside each equivalence class, and so $R_{im}(X,\tau)\subset R$ and hence $R_{im}(X,\tau)=R$, as desired. QED

  • $\begingroup$ I assume you want your pre-order to be linear (or "total"). $\endgroup$ – Goldstern Aug 12 '15 at 16:02
  • $\begingroup$ Yes, that is what I had meant. I have edited. Thanks! $\endgroup$ – Joel David Hamkins Aug 12 '15 at 16:05
  • $\begingroup$ I had thought that the method might generalize to well-quasi-orders in the infinite case, but I am confused about it. $\endgroup$ – Joel David Hamkins Aug 12 '15 at 16:48
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    $\begingroup$ @Goldstern: The problem is that the induced map on the corresponding well-ordered set does not have to be strictly order preserving, so for instance if the order-type is $\omega$ you can have the map $n\mapsto \max(0,n-1)$. This is ruled out in the finite case by the surjectivity condition. $\endgroup$ – Eric Wofsey Aug 12 '15 at 23:17
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    $\begingroup$ Yes, Eric's example is exactly the kind of thing I found to be an obstacle to my argument in the infinite case (and I was trying to do exactly what Goldstern suggested, including looking also at down-sets instead of up-sets). What I can't see is whether it truly is an obstacle or not. For example, can we have two classes, both infinite? Or infinitely many pairs? I don't know yet. $\endgroup$ – Joel David Hamkins Aug 13 '15 at 0:14

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