# What sets of self-maps are the continuous self-maps under some topology?

An open question on MSE, http://math.stackexchange.com/questions/427634/a-topology-such-that-the-continuous-functions-are-exactly-the-polynomials , asks whether there is an infinite field and a topology over that field such that the continuous functions from the field to itself are precisely the polynomials (and whether such a topology exists for $\Bbb Q$, $\Bbb R$, or $\Bbb C$). Another question, linked from there, asks the same about holomorphic functions. The common notion is this:

Given a set $X$ and a set $\mathscr C$ of functions from $X$ to $X$, what are necessary and/or sufficient conditions under which there is a topology $\mathscr T$ on $X$ such that a function $f\colon X\to X$ is $\mathscr T$-$\mathscr T$ continuous iff $f\in \mathscr C$.

It is immediately clear that $\mathscr C$ must contain the identity function and all constant functions, and must be closed under composition. Nothing else seems terribly obvious. What is known about this?

Edit: for the purpose of answering the specific questions on MSE, even results limited to connected, bihomogeneous $T_1$ spaces would be very helpful.

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Also discussed on math.SE math.stackexchange.com/questions/329545/… – Martin Brandenburg Jul 8 '13 at 18:58
@MartinBrandenburg: unfortunately there isn't much discussion there. – dfeuer Jul 8 '13 at 19:13
Magill, K. D., Jr. A survey of semigroups of continuous selfmaps. Semigroup Forum 11 (1975/76), no. 3, 189–282 and its addendum might be good first places to start. – Benjamin Steinberg Jul 8 '13 at 19:38
Another thing about the polynomials over infinite fields, and I believe also the holomorphic functions: in those cases $|X|=|\mathscr C|$. – dfeuer Jul 9 '13 at 2:41

## 3 Answers

Added.Let me explain better how this answer relates to the question since the OP asked for clarification. Being the monoid of all continuous maps is an algebraic property in the following sense. M is a monoid of mappings on a set X containing all constant maps iff M has a minimal ideal consisting of left zeroes and M acts faithfully on the left of this ideal.

So the question is equivalent to asking which abstract monoids are endomorphism monoids of topological spaces. (The set X, if it exists, must be the ideal of left zeroes.) This seems a hard question and I couldn't find an answer in the literature.

De Groot proved every group is the homeomorphism group of a topological space (and the isometry group of a metric space). Apparently he asked which monoids can be the nonconstant mappings of a topological space.

Original answer

Here is a partial answer to the second question. Among other things in Trnková, Věra Topological spaces with prescribed nonconstant continuous mappings. Trans. Amer. Math. Soc. 261 (1980), no. 2, 463–482 it is shown that given any monoid M, there is a topological space X such that the non-constant mappings on X form a monoid isomorphic to M.

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Could someone explain how this result is more than superficially related to the question? Am I missing something important? – dfeuer Jul 9 '13 at 2:15
Well it says that the monoid of cts functions on a space can be extremely general. Take any monoid you like, no matter how ugly it is, and there is a space whose endomorphism monoid is this guy and the constants. So there is no real hope to say anything nice about what the semigroup of cts mappings on a space is like from a purely algebraic point of view. – Benjamin Steinberg Jul 9 '13 at 2:51
@dfeuer, I put in a second answer since my first answer is perhaps dual in some sense to your question. – Benjamin Steinberg Jul 9 '13 at 3:41

Let me point out that the properties stated at the end are not sufficient.

Observation. There is a set $X$ with a set $F$ of functions on $X$ that contains the identity function and all constant functions and is closed under composition, but is not the set of continuous functions with respect to any topology on $X$.

Proof. Let $X=\{1, 2, 3, 4\}$, and let $F$ have the identity function, the constant functions and the permutations $(1 2)$, $(3 4)$ and $(1 2)(3 4)$. This class of functions has the identity and constant functions and is closed under composition. But I claim it is not the set of continuous functions on $X$ for any topology.

Suppose towards contradiction that $\tau$ is a topology on $X$ with $F$ being the set of all continuous functions. Note that the permutations in $F$ are self-inverse, and so they are actually homeomorphisms with respect to $\tau$. It follows that if either $1$ or $2$ is isolated, then so is the other, and the same with $3$ and $4$. Because the function $({{1234}\atop{1134}})$ is not in $F$, it follows that $\tau$ must have an open set $U$ with $1\in U$ and $2\notin U$. Similarly, there is an open set $V$ with $3\in V$ and $4\notin V$.

Let us now consider various cases. If either $U$ or $V$ is a doubleton, like $\{1,3\}$, then it easily follows that $\{1,4\}$, $\{2,3\}$ and $\{2,4\}$ are also open, and so $\tau$ is discrete, contrary to the fact that $F$ does not contain all functions on $X$. If both $U$ are tripletons, then $U\cap V=\{1,3\}$ is a doubleton, and we are in the previous case. So one of them must be a singleton. Assume without loss that $U=\{1\}$ is a singleton, and so both $1$ and $2$ are isolated. Since $\tau$ is not discrete, it follows that $3$ and $4$ are not isolated, and so we know that $\{1,2,3\}$ and $\{1,2,4\}$ are open and any open set containing either $3$ or $4$ contains both $1$ and $2$. In this case, the function $({{1234}\atop{1233}})$ would be continuous, but it is not in $F$. So there is no such topology $\tau$. QED

Similar examples can be constructed from disjoint unions of spaces, where one doesn't allow all piecewise constant functions into $F$.

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I think the number of subsemigroups of the full transformation semigroup on n labeled points exceeds by far the number of topologies on the same sized set. There should be many such examples without a topology. – The Masked Avenger Jul 9 '13 at 4:38
I'm not surprised that the properties at the end are not sufficient; they're extremely weak. I just haven't been able to come up with others! – dfeuer Jul 9 '13 at 6:44

Joel's answer is a special case of the following. A variation on Joel's answer is the following. Let G be a transitive permutation group of continuous maps on a finite topological space X with more than 2 elements. Then G together with all constants is never the whole monoid of all continuous self maps of X.

Proof. A finite topological space is just a finite preordered set via the specialization ordering/Alexandrov topology. Continuous maps are precisely order preserving maps.

If the preorder is the universal equivalence relation we have the indiscrete topology and so all maps are continuous. If the preorder is equality then the topology is discrete and so all maps are continuous. Since there are always maps on a three or more element set which are neither constant nor permutations we are done in these two cases. Let's prove only these two cases occur.

Suppose that $x\leq y$ and $gx=y$ with $g\in G$ by transitivity. Then $x\leq gx$ and so $x\leq gx\leq g^2x\leq \cdots$. Since G is finite we eventually get $g^n=1$ and so $y=gx\leq x$. Thus any comparable elements are equivalent. Hence the preorder is an equivalence relation. If there is more than 1 class and also some class is not a singleton, then crushing the non-singleton is continuous but not in the monoid. So the preorder is equality or the universal equivalence relation.

Added. The above argument shows that if G acting on a finite set X with more than 2 elements is a primitive permutation group (meaning there is no equivalence relation on X preserved by G except equality and the universal relation) then G can only be the homeomorphism group of a topology on $X$ if $G$ is $S_X$ and the topology on X is discrete or indiscrete. Thus any proper submonoid of self-maps on X whose group of units is primitive is not the monoid of all continuous mappings for any topology on X. This provides another family of examples.

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Thanks, Benjamin. But is my example really a special case of this? After all, my group of permutations is not transitive, and I had thought of that as a key reason why it worked. – Joel David Hamkins Jul 9 '13 at 3:39
@Joel, oops I read your example too quickly but it is the same issue. Each orbit of the group must have either the discrete or indiscrete topology. So you can crush 1 orbit and leave the other alone. – Benjamin Steinberg Jul 9 '13 at 3:45
Yes, that was the guiding idea of my example. – Joel David Hamkins Jul 9 '13 at 3:47
@Joel, more bad examples can be constructed by taking primitive permutation groups. See my addition. – Benjamin Steinberg Jul 9 '13 at 4:29