1
$\begingroup$

It is well known that if $X,Y$ are $T_0$ Alexandrov spaces then they are just posets. With every such spaces we can associate an abstract simplicial complex $K(X)$ where the simplices are nonempty chains in $X$. A map $f:X\to Y$ is continuous if and only if it is order preserving (as a map of posets). Hence every map $f:X\to Y$ defines a simplicial map $K(f):K(X)\to K(Y)$.

It also well know that if $X$ is finite, then if $f,g:X\to Y$ are homotopic, then so are $|K(f)|$ and $|K(g)|$.

So may question is: Is the same true when $X$ is not finite? Or maybe it is true in some particular cases. For instance when $X$ is locally finite?

It seems to be not so hard to prove or disprove but I've been trying to do this for almost a week and still I can't. Thank for any help.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Yes, this is true for arbitrary $T_0$ Alexandrov spaces. There is a natural transformation $p:|K(X)|\to X$ which is a weak equivalence by a theorem of McCord (Theorem 2 of this paper); explicitly, $p$ sends a point in the interior of a simplex corresponding to a chain in $X$ to the least element of the chain (or the greatest element, depending on your convention on which way the ordering goes). It follows that if $f,g:X\to Y$ are homotopic, then $|K(f)|$ and $|K(g)|$ are weakly homotopic (i.e., they become equal when all weak equivalences are inverted). Since $|K(X)|$ is a CW-complex, it follows that they are actually homotopic.

It would still be interesting to see if this can be shown directly without invoking McCord's theorem, which is a bit of a sledgehammer.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Could you explain what do you do you mean by ,,$|K(f)|$ and $|K(g)|$ are wekaly homotopic? It seems to me as if it was connected with model categrories, am I right? But I do not understand very well this stuff yet. $\endgroup$
    – pw1822
    Commented Feb 15, 2015 at 1:05
  • $\begingroup$ Another way to say this without using any model category machinery is that $|K(f)|$ and $|K(g)|$ become homotopic after composing with the map $|K(Y)|\to Y$. But this latter map is a weak equivalence and hence induces a bijection on homotopy classes of maps out of any CW-complex, hence $|K(f)|$ and $|K(g)|$ must be homotopic. $\endgroup$
    – Eric Wofsey
    Commented Feb 15, 2015 at 2:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .