How do simplicial maps induce maps between barycentric subdivisions?

[malicx]

Homework Statement

This is from Armstrong's Basic Toplogy, p131. It's about simplicial maps, and barycentric subdivisions where m is the m-th barycentric subdivision (defined inductively by K^m
= (K^m-1)¹

Show that a simplicial map s: |K| --> |L| induces a simplicial map from |K^m| to |L^m| for any m.

Homework Equations

The Attempt at a Solution

I've got a theorem from page 126 that says |K¹| = |K|. I just want to say that since the complexes and their subdivisions are equal, then we should have |K^m| = |K| by induction, and the same for |L|, so we have the given map. This seems far too simple, but I am not exactly sure where else to start.

 

[malicx]

bump :)

i'm still stuck at this point! does anyone have any advice? Perhaps I should just say that since s is simplicial, it takes simplexes linearly to L from K, and since K^M contains K, and L^M contains L, we have the obvious map.

 

[rasmhop]

The problem with your line of though is that yes we have an induced map,
$$f : |K^m| \to |L^m|$$
but you still need to verify that it is a simplicial map. In other words you need to confirm that it takes simplexes of $K^m$ linearly onto simplexes of $L^m$ (or if it doesn't then find a different way to induce a map that does).

Of course by the inductive definition of $K^m$ it suffices to show that you get an induced simplicial map
$$f : |K^1| \to |L^1|$$
It is not obvious that this is simplicial since K^1 has simplexes not in K.