simplicial approximation and infinite complexes It is well known that if $X$ is a finite simplicial complex then for every continuous map $f:|X|\to |Y|$ there exists a simplicial map $F: X^{(n)}\to Y$ that $|F|$ is homotopic to $f$.
Does anyone know an example of a map $f$ where $X$ is infinite and $f$ cannot be homotopic with geometric realization of any map $F: X^{(n)}\to Y$ and any $n\in \mathbb N$.
Or maybe there are some generalizations of the simplicial approximation theorem or there are known theorems that say when a map $f:X\to Y$ can be approximated by a simplicial map $F:X^{(n)}\to Y$? For example for locally finite complexes?
 A: I assume $X^{(n)}$ means the $n$th barycentric subdivision of $X$. Probably
the simplest way of demonstrating how this fails for locally finite
complexes is by a homeomorphism of 1-manifolds.
For $n \ge 3$ let $C_n$ be a triangulation of $S^1$ into $n$ vertices and
$n$ edges. Clearly $|C_n|$ and $|C_m|$ are homeomorphic and since a
homeomorphism between them has a nonzero degree, any map homotopic to
such a homeomorphism is surjective. On the other hand, for $n < m$ there
can be no simplicial mapping of $C_n$ onto $C_m$, so in that case
a homeomorphism is not homotopic to a simplicial map.
If we let $X = \coprod_{n=3}^\infty C_3$ and $Y = \coprod_{n=3}^\infty C_n$, we obviously have a homeomorphism $f: |X| \to |Y|$ and any map homotopic to $f$ must map one component of $|X|$ onto each component of $|Y|$.
However, $X^{(k)} = \coprod_{n=3}^\infty C_{3\cdot 2^k}$, so there is no
component of $X^{(k)}$ that can be mapped simplicially onto $C_{3\cdot 2^k+1} \subset Y$.
Thus barycentric subdivision fails to produce a simplicial approximation for maps of locally finite complexes. It does not seem impossible to devise a more flexible strategy that could have different levels of
subdivision in different parts of the complex, but working out the details would probably be a lot of work and it may be easier to turn to cellular
approximation instead.
