Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It is a basic topological fact that CW-complexes aren't typically metrizable (they must satisfy a certain local finiteness condition) and the quotient topology is to blame.

Suppose $X$ is a finite dimensional CW-complex with countably many cells in each dimension. Is it possible weaken the topology of $X$, without changing homotopy type, so the resulting space is metrizable?

share|improve this question
3  
A classical result of Whitehead says: every countable finite dimensional CW complex is homotopy equivalent to a locally finite CW complex of the same dimension, which is of course metrizable. –  Igor Belegradek Jan 13 '13 at 19:49
1  
Thank you for these comments @Igor and @Misha. I am aware of Whitehead's result, however, I don't want to lose my preferred CW-structure with a non-explicit homotopy equivalence. What I am asking seems plausible when you consider the 1-dimensional case. For instance, a countably infinite wedge of circles is not first countable but you can weaken the topology at the basepoint so that it embeds in $\mathbb{R}^2$. The homotopy inverse of the continuous (but non-open) identity map comes from collapsing a small closed ball about the basepoint. –  Jeremy Brazas Jan 14 '13 at 15:21
1  
Jeremy, actually the homotopy equivalence can be described explicitly (increasing dimension by one), i.e. the locally finite complex can be obtained as a mapping telescope of an exhausion of your given countable complex by finite subcomplexes (as explained somewhere in Hatcher's "Algebraic topology" text). Maybe this is be enough for your purposes. –  Igor Belegradek Jan 14 '13 at 21:48
1  
Jeremy: The explicit construction Igor refers to is on the last page of math.cornell.edu/~hatcher/AT/AT-exercises.pdf –  Misha Jan 15 '13 at 0:09
2  
I did not mean to stress the "non-explicit homotopy" as much as "preferred CW-structure" but I will see if I can make the construction work for me. Regardless, I still hope someone might know the answer to my question. Thanks again! –  Jeremy Brazas Jan 15 '13 at 14:30

1 Answer 1

This addresses the modified question in Jeremy's comments, on keeping the preferred CW-structure.

1) If the CW complex happens to be regular and PL (i.e. the attaching maps are injective and piecewise-linear), its barycentric subdivision is a simplicial complex (namely, the order complex of the poset of nonempty faces of the CW complex), which can be endowed with the usual barycentric metric. The identity map will then be a homotopy equivalence (proofs can be found in some old textbooks, including the Appendix of Dold's Algebraic topology, or "Theory of retracts" by Hu Sze-Tsen).

2) For a general (countable) CW complex, one can inductively homotop the attaching maps of $(n+1)$-cells by a homotopy with values in the $n$-skeleton so that the modified CW complex $K$ admits a barycentric subdivision $K'$ that is a regular simplicial set, in the sense that cells of $K$ are identified with the unions of simplices of $K'$ whose first vertex is a fixed $0$-simplex of $K'$. (Regular means that the representing map of every non-degenerate simplex only makes identifications along the last facet of the simplex.) The geometric realization of a regular simplicial set is a regular CW-complex, so the previous construction applies. (In more detail, the order complex $K''$ of the poset of nonempty nondegenerage simplices of $K'$, ordered by inclusion, is a simplicial complex.) An enlightening overview of subdivisions of simplicial sets can be found here.

The homotopies of attaching maps can be constructed using Brouwer's simplicial approximation theorem, which implies that any continuous map $|L|\to |X|$ between geometric realizations of finite simplicial sets is homotopic, upon precomposing with the geometric realization of an iterate $L^{(n)}\to L$ of the last vertex map $L'\to L$, to the geometric realization of a morphism $f:L^{(n)}\to X$ of simplicial sets (see Corollary 3.2 here). Here $L$ is any triangulation $S^n$ by a non-singular simplicial set (i.e. a subcomplex of the order complex of a poset), and $X$ is the $n$-skeleton of $K'$, which is a regular simplicial set. Then the mapping cone of $f$ is again a regular simplical set.

In light of the combinatorial view of regular PL CW complexes, one could try to homotop the attaching maps of a general CW complex so as to achieve a more rigid combinatorial structure of the simplicial set $K'$. However, a homotopy class is not generally representable by a non-degenerate map (in the sense of collapsing no simplices). Because of this, $K'$ cannot be generally chosen to be the nerve of a category, nor even a quasi-category.

3) One drawback of the metric topology on simplicial complexes is that it indeed is incompatible with quotients (they are non-metrizable, unless each equivalence class is compact). This difficulty can be avoided by endowing the simplicial complexes with a "cubical" l_infty metric and working uniformly. This applies to regular PL CW complexes, as well as to those CW complexes whose attaching maps are jointly uniformly continuous (using iterates of canonical, rather than barycentric, subdivision and Theorem 7.4 here).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.