# What spaces $X$ do have $\text{End}(X) \cong \text{End}(\mathbb{R})$?

This is a follow-up on the following question. Let $$\text{End}(X)$$ denote the endomorphism monoid of a topological space $$X$$ (that is, the collection of all continuous maps $$f:X\to X$$ with composition).

What is an example of a topological space $$X$$ with $$X\not\cong \mathbb{R}$$ but the monoids $$\text{End}(X)$$ and $$\text{End}(\mathbb{R})$$ are isomorphic?

• A general remark: in $\mathrm{End}(X)$, constants are precisely the left-absorbing elements (i.e., those $g$ such that $gf=g$ for all $f$). The monoid acts on the left on left-absorbing elements, and this is conjugate to the action on $X$. In particular, the condition that $X$ is homogeneous under $\mathrm{Aut}(X)$ can be read on the abstract monoid $\mathrm{End}(X)$. I expect this is might be a starting point to check that every $X$ with $\mathrm{End}(X)\simeq\mathrm{End}(\mathbf{R})$ is homeomorphic to $\mathbf{R}$. – YCor Mar 29 at 15:17
• The constant maps are not enough in general to pick up the topology for a general space. For example the endomorphism monoid of X in the discrete and indiscrete topology are the same. – Benjamin Steinberg Mar 29 at 16:55
• Indeed it sounds tricky, even assuming beforehand that $X$ is Hausdorff. – YCor Mar 29 at 17:57

No such space exists. We actually get the stronger statement that every isomorphism $$\operatorname{End}(X) \stackrel\sim\to \operatorname{End}(\mathbf R)$$ is induced by an isomorphism $$X \stackrel\sim\to \mathbf R$$ (unique by Observation 1 below). In contrast, in Emil Jeřábek's beautiful construction in this parallel post there is an 'outer automorphism' $$\operatorname{End}(X) \stackrel\sim\to \operatorname{End}(X)$$ that does not come from an automorphism $$X \stackrel\sim\to X$$ of topological spaces (it comes from an anti-automorphism of ordered sets).

I will use the substantial progress by YCor and Johannes Hahn, summarised as follows:

Observation 1 (YCor). For every topological space $$X$$, the map $$X \to \operatorname{End}(X)$$ taking $$x$$ to the constant function $$f_x$$ with value $$x$$ identifies $$X$$ with the set of left absorbing¹ elements of $$\operatorname{End}(X)$$.

In particular, an isomorphism of monoids $$\operatorname{End}(X) \stackrel\sim\to \operatorname{End}(Y)$$ induces a bijection $$U(X) \stackrel\sim\to U(Y)$$ on the underlying sets.

Observation 2 (Johannes Hahn). If $$\operatorname{End}(X) \cong \operatorname{End}(\mathbf R)$$, then $$X$$ is $$T_1$$. Since the closed subsets of $$\mathbf R$$ are exactly the sets of the form $$f^{-1}(x)$$ for $$x \in \mathbf R$$, we conclude that these are closed in $$X$$ as well, so the bijection $$X \to \mathbf R$$ of Observation 1 is continuous.

(The asymmetry is because we used specific knowledge about $$\mathbf R$$ that we do not have about $$X$$.)

To conclude, we prove the following lemma.

Lemma. Let $$\mathcal T$$ be the standard topology on $$\mathbf R$$, and let $$\mathcal T' \supsetneq \mathcal T$$ be a strictly finer topology. If all continuous maps $$f \colon \mathbf R \to \mathbf R$$ for $$\mathcal T$$ are continuous for $$\mathcal T'$$, then $$\mathcal T'$$ is the discrete topology.

Note that Observation 2 and the assumption $$\operatorname{End}(X) \cong \operatorname{End}(\mathbf R)$$ imply the hypotheses of the lemma, so we conclude that either $$X = \mathbf R$$ or $$X = \mathbf R^{\operatorname{disc}}$$. The latter is clearly impossible as it has many more continuous self-maps.

Proof of Lemma. Let $$U \subseteq \mathbf R$$ be an open set for $$\mathcal T'$$ which is not open for $$\mathcal T$$. Then there exists a point $$x \in U$$ such that for all $$n \in \mathbf N$$ there exists $$x_n \in \mathbf R$$ with $$|x - x_n| \leq 2^{-n}$$ and $$x_n \not\in U$$. Without loss of generality, infinitely many $$x_n$$ are greater than $$x$$, and we can throw out the ones that aren't (shifting all the labels, so that $$x_0 > x_1 > \ldots > x$$). Up to an automorphism of $$\mathbf R$$, we can assume $$x = 0$$ and $$x_n = 2^{-n}$$ for all $$n \in \mathbf N$$. Taking the union of $$U$$ with the usual opens $$(-\infty,0)$$, $$(1,\infty)$$, and $$(2^{-n},2^{-n+1})$$ for all $$n \in \mathbf N$$ shows that $$Z = \big\{1,\tfrac{1}{2},\tfrac{1}{4},\ldots\big\}$$ is closed for $$\mathcal T'$$. Consider the continuous function \begin{align*} f \colon \mathbf R &\to \mathbf R\\ x &\mapsto \begin{cases}0, & x \leq 0,\\ x, & x \geq 1, \\ 2^nx, & x \in \big(2^{-2n},2^{-2n+1}\big], \\ 2^{-n}, & x \in \big(2^{-2n-1},2^{-2n}\big].\end{cases} \end{align*} Then $$f^{-1}(Z)$$ is the countable union of closed intervals $$Z' = \bigcup_{n \in \mathbf N} \big[2^{-2n-1},2^{-2n}\big] = \big[\tfrac{1}{2},1\big] \cup \big[\tfrac{1}{8},\tfrac{1}{4}\big] \cup \ldots.$$ By the assumption of the lemma, both $$Z'$$ and $$2Z'$$ are closed in $$\mathcal T'$$, hence so is the union $$Z'' = Z' \cup 2Z' \cup [2,\infty) = (0,\infty),$$ and finally so is $$Z'' \cup (-Z'') = \mathbf R\setminus 0$$. Thus $$0$$ is open in $$\mathcal T'$$, hence so is every point, so $$\mathcal T'$$ is the discrete topology. $$\square$$

¹Elements $$f$$ such that $$fg = f$$ for all $$g$$. (I would probably have called this right absorbing!)

• I checked before posting: the condition that $f$ satisfies $\forall g,fg=g$ is always called "left-absorbing", or "left zero" (which sounds intuitive to me: $f$ absorbs on the lefts elements from the right). – YCor Mar 30 at 14:34
• @YCor: sorry, I wasn't trying to challenge your use. I also checked and saw that this is what it's called. Anyway, left and right are hard. – R. van Dobben de Bruyn Mar 30 at 22:12
• No problem. In the same vein, in a group $G$ with subgroup $H$, one calls $gH$ left coset and $Hg$ right coset but this is possibly not obvious either. – YCor Mar 30 at 22:23
• I feel that YCor's guys are "left absorbing" or "absorbing right" (in that order). – Wlod AA Mar 31 at 6:40
• Over in the other thread I said "left-absorbing", but it means of course absorbing to the left! – Todd Trimble Mar 31 at 10:18

By @YCor's comment, $$X$$ has the same number of elements as $$\mathbb{R}$$ and the actions of $$End(X)$$ on $$X$$ is the same as the action of $$End(\mathbb{R})$$ on $$\mathbb{R}$$. Now consider the automorphism group and its action on $$X$$. $$Aut(\mathbb{R})$$ consists of strictly increasing and strictly decreasing maps and it is easy to see that the point stabiliser subgroups $$Aut(\mathbb{R})_x$$ act transitively on $$\mathbb{R}\setminus\{x\}$$. But this action is imprimitive: $$\mathbb{R}\setminus\{x\}$$ has two blocks, namely $$(x,+\infty)$$ and $$(-\infty,x)$$ and the set stabilisers act transitively on those two sets. Phrased differently: There are exactly three equivalence relations on $$\mathbb{R}\setminus\{x\}$$ invariant under $$Aut(\mathbb{R})_x$$, the two trivial ones "everything is equivalent" and "nothing is equivalent" as well as a unique non-trivial one with the two equivalence classes $$(-\infty,x)$$ and $$(x,+\infty)$$.

Conclusion: We can recover a linear order on $$X$$ from $$Aut(X)$$ and thus from $$End(X)$$. And in particular we can recover the order topology of this ordering. And all elements of $$End(X)$$ must be continuous w.r.t. this order topology.

This does not necessarily mean that the order topology coincides with the original topology on $$X$$, but it is very close.

What can we say about the original topology? We know that there are lots of continuous map, but not too many (since $$|End(\mathbb{R})|=|\mathbb{R}|$$) so that $$X$$ is not indiscrete. We can say that the only self-maps with finite image are constant. In particular, there cannot be a Sierpinski space inside $$X$$, because every open set yields a continuous map to the Sierpinski space. Therefore $$X$$ is at least a $$T_1$$-space.

That is enough to conclude that the original topology on $$X$$ is at least as fine as the order topology: For every order-topology-closed subset $$A\subseteq\mathbb{R}$$ there is a continuous self-map $$f:\mathbb{R}\to\mathbb{R}$$ with $$A=f^{-1}(0)$$. Therefore the corresponding order-topology-closed subset of $$X$$ is also the preimage of a point under a continuous self-map and thus original-topology-closed. In other words: Our canonical bijection $$X\to\mathbb{R}$$ is continuous w.r.t. the original topology on $$X$$.

I have the feeling that the other direction is equally easy, but I just don't see it right now.

• If you have a continuous bijection onto $\mathbf{R}$, you have the Hausdorff property. I've been wondering if there are strange strong topologies (such as the one generated by the standard topology and the co-countable subsets), maybe the latter has the same homeomorphism group, but not the same monoid of continuous self-maps). – YCor Mar 29 at 18:26
• Thanks for this effort that is a key ingredient towards answering the question! – Dominic van der Zypen Mar 30 at 8:45

There is a straightforward approach to proving that isomorphism of monoids $$\ \text{End}(X\ T)\$$ and $$\ \text{End}(\Bbb R\,\ T_E)\$$ implies homeomorphism of $$\ (X\ T)\$$ and Euclidean space $$\ (\Bbb R\,\ T_E):\$$

let $$\ (M\ \circ\ J)\$$ be an arbitrary abstract monoid. One may associate with this monoid a set of "points" $$\ C\$$ of "constant" elements, as @YCor did in his first comment, as the left-absorbing elements $$\ c\in C,\$$ where

$$\forall_{f\in M}\quad c\circ f=c$$

Then one selects purely algebraic (monoidal) notions which have the respective topological meaning in $$\ (\Bbb R\,\ T_E).\$$ You need just a couple of such notions.

Then you force on the abstract monoid $$\ (M\ \circ\ J)\$$ the purely algebraic (monoidal) axioms that induce the properties of reals. That's all.

For instance one may apply idempotents $$\ i\in\mathcal I\subseteq M,\$$ where $$\ i\circ i=i. \$$ Then the critical notion to me is what I've defined and call universally closed idempotents or uc-morphism $$\ i\in\mathcal I,\$$ which satisfy:

$$\forall_{f\in M}\,\exists_{p\in M}\quad i\circ f\circ i\circ p\,\ =\,\ i\circ p$$

Observe that we have a canonical map

$$\pi: \mathcal I\to 2^C,$$

where

$$\forall_{i\in\mathcal I}\quad \pi(i)\ := \ \{c\in C:\ i\circ c=c\}$$

This has nice properties... etc.

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(After I saw the OP question, I hesitated... and I can remove my post, no sweat.)

• Also, idempotents that are not universally closed (or nuc-morphisms for short) have their role as well. – Wlod AA Mar 31 at 6:47
• Dominic shoots from the hip on MO but this OP-question and his other one, EndMonDom-question are well aimed! The latter offers great scope and allows for a general presentation of the topic. (I saw this latter but earlier question only now). – Wlod AA Mar 31 at 14:35