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If $(X,\tau)$ is a topological space and $x,y\in X$ we say that $x$ is mappable to $y$ if there is a non-constant continuous map $f:X\to X$ with $f(x) = y$.

Is there a Hausdorff space $X$ with more than one point such that whenever $x\neq y\in X$ and $x$ is mappable to $y$, then $y$ is not mappable to $x$?

Dominic van der Zypen
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  • Do you mean a non-constant continuous map? –  Feb 02 '17 at 08:09
  • mappable?--how about "$x$ pays attention to $y$" or "$x$ appreciates $y$" or .... – Włodzimierz Holsztyński Feb 02 '17 at 08:14
  • @Anton, in topology it's common to assume that maps or mappings stand for continuous functions. – Włodzimierz Holsztyński Feb 02 '17 at 08:17
  • The question seems about quasi-ordering not being a partial ordering(?). – Włodzimierz Holsztyński Feb 02 '17 at 08:22
  • Is there a Hausdorff space (with more than one point) having no non-constant non-identity continuous self-maps? So every continuous self-map is either constant or the identity. Such a space would be an example. – Joel David Hamkins Feb 02 '17 at 14:09
  • @JoelDavidHamkins: Good point. And yes, there are such spaces, like the one Ramiro de la Vega points to in this related question: http://mathoverflow.net/questions/188729/strongly-rigid-hausdorff-spaces?rq=1 – Will Brian Feb 02 '17 at 14:46
  • I thought that there might be such a space amongst the continua. And I had seen Cook's theorem, but somehow thought it was only about homeomorphisms. Does "map" for Cook mean continuous function only? – Joel David Hamkins Feb 02 '17 at 14:56
  • Cook claims to answer my question in his preamble, but I am confused about whether his theorem statement (theorem 9) actually does answer it. – Joel David Hamkins Feb 02 '17 at 15:01
  • @JoelDavidHamkins: I just looked at the paper. I think Theorem 11 is the one that clearly answers the question (it says: the identity is the only mapping of M_2 onto a non-degenerate subcontinuum of M_2). Here "mapping" means continuous function and "non-degenerate subcontinuum" means a closed, connected subspace with more than one point. Since continuous functions on M_2 always have closed, connected images, we can interpret this theorem to say that any continuous function from M_2 to itself, other than the identity, is constant. – Will Brian Feb 02 '17 at 15:21
  • Great! Why don't you summarize it all in an answer? This space has the desired property vacuously. – Joel David Hamkins Feb 02 '17 at 15:26
  • I feel that the question was somewhat incomplete. I'd add a requirement that there exist $\ x\ y\in X\ $ such that $\ x\ $ is mappable onto y. One may even consider a stronger requirement (a narrower class): every $\ x\in X\ $ is either mappable onto another point or a point different from $\ x\ $ is mappable onto $\ x$. – Włodzimierz Holsztyński Feb 02 '17 at 23:10

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Yes, there are such spaces. One example is referenced by Ramiro de la Vega in his answer to this related question.

As Joel points out in the comments, the property you describe, namely "$X$ has an anti-symmetric mappability relation", is strictly weaker than the property "$X$ is strongly rigid", where the latter is defined to mean that the only continuous functions from $X$ to itself are the identity map and the constant functions.

The paper mentioned in Ramiro's answer (available here) gives a hereditarily indecomposable continuum with this property. The construction is due to Howard Cook, and the fact that it has the desired property is expressed in Theorem 11 of his paper. (The theorem states that the identity is the only mapping of $M_2$ onto a non-degenerate subcontinuum of $M_2$ (where $M_2$ is a continuum constructed earlier). Here mapping means continuous function and non-degenerate subcontinuum means a closed connected set having more than one point. Since the continuous image of $M_2$ is always closed and connected, this theorem implies immediately that the only non-identity continuous functions from $M_2$ to itself are the constant maps.)

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