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Dec 14, 2022 at 8:57 comment added HJRW @user494312: I said simplicial complexes, which of course are always locally connected. Sorry if I was unclear: I just mean that the FPT for T_0 finite spaces can be proved this way. I have nothing to say about spectral spaces.
Dec 13, 2022 at 19:14 comment added user494312 @HJRW Don't you need properties like locally connected for good behaviour of singular complexes ?
Dec 13, 2022 at 9:33 comment added user494312 @Z. M : but actually yes, if you replace 'contractible' and/or 'absolute extensor' by a condition in terms of the maps of finite spaces, you might get a conjecture about fixed property which applies to spectral spaces. The contractability-like condition on $X$ is that $X\to\{pt\}$ lies in the orthogonal wrt the weak lifting property $f^{lr}$ where $f$ is a map of finite topological spaces which is an acyclic Serre fibration. For some $f$ and $X$ a finite CW complex it does define contractability.
Dec 13, 2022 at 9:26 comment added HJRW @Z.M: thanks for pointing that out, I hadn’t clocked that the T_0 assumption was still in force in the final paragraph. In that case, the relevant fixed point theorem should be a trivial consequence of the Lefschetz FPT for simplicial complexes. (Presumably everyone knows this, but I thought I’d mention it explicitly since no one else has yet.)
Dec 13, 2022 at 9:20 comment added user494312 @Z. M : As for now, I do not know. I am following up the references by @ Michael Greinecker, and may update the post with references and a conjecture which might apply to spectral spaces.
Dec 13, 2022 at 9:18 comment added user494312 So is there a form of fixed point theorem for spectral spaces ?
Dec 13, 2022 at 8:35 comment added Z. M By "this property", I mean the fixed point property that you want — namely, that property on spectral spaces (but that does not incorporate the version for compact Hausdorff spaces that you want). As I explained, spectral spaces are "profinite" sober spaces.
Dec 13, 2022 at 8:31 comment added Z. M @HJRW OP mentioned this in the post: the necessity of $T_0$.
Dec 12, 2022 at 17:59 comment added HJRW Without knowing anything about finite topological spaces, I'm surprised that the generalisation of the BFPT you mention is true. I would have guessed that the geometric realisation $|K|$ of the two-point space with the indiscrete topology $K$ is the closed interval. However, the involution that swaps the two points has no fixed point in $K$, but induces a homeomorphism that fixes a point in $|K|$. Apologies if I've misunderstood; I'm entirely ignorant of this subject!
Dec 12, 2022 at 17:10 comment added user494312 In fact, the modification of contractibility I'd like to use is as follows: X is such that that $X\to \{pt\}$ is in the left-then-right orthogonal of a particular map $M\to\Lambda$ of finite topological spaces which is an acyclic Serre fibration and is a model of barycentric subdivision of the interval, defined in the list here. It is somewhat similar to being an absolute extensor.
Dec 12, 2022 at 17:05 comment added user494312 Which 'this property'? I think the usual geometric realisation of a poset (viewed as a simplicial set of its chains or some such) is a retract of a filtered colimit of barycentric subdivisions, see this thesis, Theorem 2.4.14 which claims it is at least homotopy equivalent, referenced by P.May.
Dec 12, 2022 at 15:54 comment added Z. M I mean, the geometric realization of spectral spaces is defined to be the right Kan extension from finite sober spaces, so it will take value in compact Hausdorff spaces.
Dec 12, 2022 at 15:38 comment added Z. M So does it imply that spectral spaces also have this property, since spectral spaces are characterized by being cofiltered limits of finite sober (or equivalently, $T_0$-)spaces?
Dec 12, 2022 at 13:05 comment added user494312 @Z.M this is what is defined in the Clader's thesis. ( I seem to recall people writing there was an inaccuracy/error there but do not recall at the moment). Maybe also see this How much of homotopy theory can be done using only finite topological spaces?
S Dec 12, 2022 at 11:38 history suggested Z. M CC BY-SA 4.0
Fix markdown & LaTeX
Dec 12, 2022 at 10:09 comment added Z. M What is the geometric realization in the last paragraph?
Dec 12, 2022 at 10:08 review Suggested edits
S Dec 12, 2022 at 11:38
Dec 11, 2022 at 21:55 comment added Michael Greinecker I'm not that knowledgable when it comes to algebraic topology, but "acyclic absolute neighborhood retract" is enough by a fixed-point theorem of Eilenberg and Montgomery.
Dec 11, 2022 at 21:35 comment added user494312 Thanks! I replaced "contractible" with "absolute extensor for normal spaces", which is different. Would that help ? Is this counterexample an absolute extensor for normal spaces ?
Dec 11, 2022 at 21:29 history edited user494312 CC BY-SA 4.0
replaced "contractible" by "absolute extensor" in response to comment by
Dec 11, 2022 at 19:34 comment added Michael Greinecker There is a famous example by Kinoshita of a contractible compact subset of $\mathbb{R}^3$ that lacks the topological fixed-point property.
Dec 11, 2022 at 19:26 history asked user494312 CC BY-SA 4.0