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I'm surprised that I can't find any research on this topic. Maybe it's too obvious? Kinoshita proved that contractible continuum do not have FPP, but his example is not locally connected. Maybe if we add this to the conditions it will have FPP?

UPD: Continuum as a nonempty compact connected metric space

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – CrSb0001
    May 31 at 18:34
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    $\begingroup$ @CrSb0001 How can i impove it? I'm asking for answer yes or not with maybe some links on research in this topic. Question "Does a contractible locally connected continuum have an fixed point property?" not clear? $\endgroup$
    – LoliDeveloper
    May 31 at 18:40
  • $\begingroup$ The function $f(x)=x+1$ on $\Bbb{R}$ does not have a fixed point. Does that answer the question? $\endgroup$
    – John Doe
    May 31 at 19:01
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    $\begingroup$ @JohnDoe No. I mean en.wikipedia.org/wiki/Continuum_(topology) continuum as nonempty compact connected metric space. Thank you. I'll add it to question $\endgroup$
    – LoliDeveloper
    May 31 at 19:03

1 Answer 1

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While I do not have an answer to your question, here are two related results:

  1. If $X$ is a 1-dimensional contractible continuum, then $X$ satisfies the FPP (every continuous self-map $X\to X$ has a fixed point). See:

Young, G. S., Fixed-point theorems for arcwise connected continua, Proc. Am. Math. Soc. 11, 880-884 (1961). ZBL0102.37806.

  1. Suppose that $X$ is an acyclic compact ANR. Then $X$ satisfies the FPP. See Corollary 8.10 on page 68 in

Górniewicz, L., On the Lefschetz fixed point theorem, Brown, R.F. (ed.) et al., Handbook of topological fixed point theory. Berlin: Springer (ISBN 1-4020-3221-8/hbk). 43-82 (2005). ZBL1077.55001.

In order to connect to your question, note that every ANR is locally contractible and almost conversely every finite dimensional metrizable locally contractible space is an ANR.

My guess is that one should assume in your question local contractibility instead of local connectivity, but I do not have an example.

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  • $\begingroup$ Thank you. I still don't understand how to connect these results, because there is no acyclic condition. About local properties, i meant local connectivity, because Kinoshita's example of contractible continua was not localy connected. Please don't mark this answer as solve, maybe someone have more information to share. $\endgroup$
    – LoliDeveloper
    May 31 at 23:14
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    $\begingroup$ @LoliDeveloper: Did you take any algebraic topology classes? Acyclic is weaker than contractible. $\endgroup$
    – Moishe Kohan
    May 31 at 23:20
  • $\begingroup$ Oh, sure, my bad. Kinda beginner in this area $\endgroup$
    – LoliDeveloper
    May 31 at 23:28
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    $\begingroup$ It is not locally simply connected, hence, not an ANR. Of course, ANR is stronger than local connectivity. $\endgroup$
    – Moishe Kohan
    Jun 1 at 15:16
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    $\begingroup$ @LoliDeveloper: By the way, only you can accept an answer to your questions, by clicking the checkmark next to an answer. Since no true answer has emerged so far (after 10 days), my suggestion is to ask at MathOverflow, but refer to this question at Math Stack Exchange as a cross-post. It is a good question. $\endgroup$
    – Moishe Kohan
    Jun 11 at 14:43

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