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. 



homotopic maps of locally finite spaces
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](http://projecteuclid.org/euclid.dmj/1077376525)); 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.


