Does every set $X$ have a topology for which the only continuous self-surjection is the identity map?

This question is a special case of Dominic van der Zypen's question Reconstructing relations with the image relation of a topology, as discussed in the comments, particularly the comment of Eric Wofsey, which explains that if the answer to this question is affirmative, then the answer to Dominic's question is also affirmative.

Question. Does every set $X$ have a topology for which the only continuous surjection $f:X\to X$ is the identity map?

My answer to Dominic's question shows that for finite sets, the answer is affirmative, since one need only place an order on $X$, and then let the topology be the up-sets of the order. Every continuous surjection is a permutation of $X$ and order-preserving, and hence the identity. Although it is tempting to try to use the same method with well-orders on an infinite set, it doesn't quite work out in that generality, because the predecessor function on the finite height elements (otherwise fixed) will be a continuous surjection, but not the identity.

For an extreme negative answer, I would ask: can one show that there is no topology on a countably infinite set for which the only continuous surjection is the identity map?

• I have a feeling that the rigidity results mentioned in mathoverflow.net/a/6300/1946 will be relevant. – Joel David Hamkins Aug 13 '15 at 4:01
• This question also seems related. In particular, the answer there gives an example with $|X|=\mathfrak{c}$. – Eric Wofsey Aug 13 '15 at 4:22
• This answer also gives an example whenever $|X|^{\aleph_0}=|X|$. Still, it seems like this sort of thing is massive overkill and there ought to be some fairly simple construction that always works. – Eric Wofsey Aug 13 '15 at 4:50
• Very nice question! I think spaces $(X,\tau)$ with the property that the identity is the only continuous self-surjection must have some low separation property: if $x\neq y\in X$ have the property that ${\cal N}_x = {\cal N}_y$ then a continuous self-bijection that swaps $x, y$ can be constructed (if I'm not mistaken). – Dominic van der Zypen Aug 14 '15 at 7:15

For $X$ of cardinality at least continuum there exist such spaces, even metrizable ones. It follows immediately from a much more general result, proved by Trnková in [1]. She proves that the category of graphs admits a full embedding into the category of metrizable spaces with nonconstant maps. An inspection of the proof shows that a graph $G$ is sent to a space whose cardinality is $|G|\times\frak{c}$. Then we use the fact that there exist rigid graphs $G$ of every cardinality - this is proved in P. Vopěnka, A. Pultr and Z. Hedrlin, Commentationes Mathematicae Universitatis Carolinae 6(1965), 149-155.

[1] V. Trnková, Non-constant continuous mappings of metric or compact Hausdorff spaces, Comment. Math. Univ. Carolinae 13 (1972) 283–295.

• Would it be possible for you to sketch the construction? – Joel David Hamkins Aug 13 '15 at 12:27
• Note also that you use exactly the rigidity result of Vopěnka, Pultr and Hedrlin to which I had alluded in my comment on the original post! – Joel David Hamkins Aug 13 '15 at 12:28
• @Joel David Hamkins -- One takes a continuum $C$ constructed by Cook (Eric Wofsey points to it in his comment). It has the properties that any map $C\to C$ is either constant or the identity and for every continuum $H\subseteq C$ other than a point and a nonconstant map $f:H\to C$, $f$ is a retraction onto some $K\subseteq H$. Then you pick two distinct points in $C$ (call them initial and terminal) and replace every edge in $G$ with a copy of $C$. I think that this works fine except it does not prevent edges from collapsing. So she introduces two additional rigid continua... – Adam Przeździecki Aug 14 '15 at 6:21
• This is about what I got from the paper. I never read it, just skimmed for this answer. She also cites some preprocessing to reduce the problem to connected graphs without loops but I think that this is not necessary. – Adam Przeździecki Aug 14 '15 at 6:21
• If the embedding in question does not turn any non-surjection into a surjection (don't know whether this is true, but it seems plausible), then the Vopěnka-Pultr-Hedrlin result is unnecessary, and you can just use ordinals (with their strict ordering). – Eric Wofsey Aug 15 '15 at 0:32

As a very special case, I claim that there is no regular topology on a countably infinite set such that the only contiuous surjection $f:X\rightarrow X$ such that is the identity map. If $X$ is countably and regular, then $X$ is a regular Lindelof space. Since every regular Lindelof space is paracompact, the space $X$ is paracompact and hence completely regular and even realcompact. If $X$ is compact, then $X$ is isomorphic to some countable ordinal $\alpha+1$. Therefore, the mapping $g:\alpha+1\rightarrow\alpha+1$ where $g(n)=n-1$ where $n$ is a finite non-zero ordinal and $g(\beta)=\beta$ where $\beta$ is infinite or zero is a continuous surjection.

Now assume that $X$ is not compact. Recall that a space is compact if and only if it is realcompact and pseudocompact. Therefore, since $X$ is not compact but $X$ is realcompact, the space $X$ is not pseudocompact. Furthermore, the space $X$ is zero-dimensional by the following argument: if $U$ is a neighborhood of $x$ and $f:X\rightarrow[0,1]$ is a mapping such that $f(x)=1$ and $f=0$ outside $U$, then the function $f$ is not surjective since $X$ is countable. Therefore, if $r\not\in f[X]$, then $f^{-1}[r,1]$ is a clopen set with $x\in f^{-1}[r,1]\subseteq U$. Therefore, since $X$ is zero-dimensional but not pseudocompact, by this answer, there is a partition of $X$ into infinitely many clopen sets $(C_{n})_{n\in\mathbb{N}}$. Let $(x_{n})_{n\in\mathbb{N}}$ be an enumeration of the elements of $X$. Then let $f:X\rightarrow X$ be the mapping where we let $f(x)=x_{n}$ whenever $x\in C_{n}$. Then $f$ is generally a non-identity continuous surjection.

I conjecture that it is consistent with the negation of the continuum hypothesis that every completely regular space of cardinality below the continuum has a non-identity surjection.

In the article Constructions and Applications of Rigid Spaces, I (Advances in Mathematics 29, 89--130 (1978), V. Kannan and M. Rajagopalan show that if $(2^{\aleph_0})^+ < 2^{2^{\aleph_0}}$, there is a countable space $X$ such that the only non-constant continous self-map is the identity. (Of course that's a notion of rigidity that is "more rigid" than what is asked in the original post.)