Making CW-complexes metrizable

Question

$\newcommand\met{\mathrm{met}}$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.

Question: Suppose $X$ is a CW-complex (possibly with countably many cells and maybe even of finite dimension). Is it possible weaken the topology of $X$ to construct another space $X_{\met}$ (with the same underlying set), so that the continuous identity function $X\to X_{\met}$ is a homotopy equivalence?

Update: I will clarify (now much later) that this question has an affirmative answer for simplicial complexes. Given an arbitrary simplicial complex $K$, we have $|K|$, which has the weak topology and is not always metrizable. However, you can give the underlying set of $|K|$ a metrizable topology to form the "metric simplicial complex" $|K|_m$. The identity $|K|\to |K|_m$ is continuous and is a homotopy equivalence. A nice proof can be found in Segal and Mardesic's book Shape Theory in Appendix, $\S 1.3$, Theorem 10. As Sergey Melikhov nicely points out in his answer, the same is true for regular CW-complexes, which include simplicial complexes. Using this result, it follows that every CW-complex is homotopy equivalent to some metric space. However, my question is a bit more specific.

Answer 1

The required metric topology on $X$ does exist. This is a consequence of Theorem 2.1 in the paper

Robert Cauty, Rétractions dans les espaces stratifiables, Bulletin de la Société Mathématique de France, 102, (1974), 129-149.

Actually, Cauty's statement is far more general than what is being required. I'll record it in the present context and postpone discussion for afterwards.

Theorem (Cauty): Let $X$ be a CW complex. Then there exists a continuous metric $d$ on $X$ such that;

• $X_{Met}=(X,d)$ is an ANR.

• The identity $i:X\rightarrow X_{Met}$ is a homotopy equivalence.

• For any given compact $K\subseteq X$, a homotopy inverse $j:X_{Met}\rightarrow X$ to $i$ can be found so that $i\circ j\simeq id_{X_{met}}$ and $j\circ i\simeq id_X$ by homotopies fixing $K$ pointwise at all times. $\quad\blacksquare$

There are no restrictions on the dimension of $X$ or its number of cells. A metric is continuous if and only if it comes from a weaker metric topology.

Now, as mentioned above, Cauty's actual statement is far more general.

The theorem holds verbatim when $X$ is any ANR for stratifiable spaces.

The term stratifiable is meant in the sense of Borges, who established already that a stratifiable space belongs to ANR(stratifiable) if and only if it belongs to ANE(stratifiable). Ceder had previously established the result that all CW complexes are stratifiable, and Cauty subsequently showed that all CW complexes belong to ANR(stratifiable).

Ignoring all references to stratifiability, we can at the very least use the extra generality to extend the theorem stated above to include the cases $(i)$ $X$ is the product of finitely many CW complexes, $(ii)$ $X$ is an open subset of a CW complex, $(iii)$ $X$ is a closed subset of a CW complex $Y$ whose inclusion $X\subseteq Y$ is a cofibration, $(iv)$ $X$ is the loop space of a CW complex.

Finally, let us comment on the last listed property regarding the compact $K$. Any compact subset $K\subseteq X$ has the same topology as its image in $X_{Met}$. The statement says that $i:X\rightarrow X_{Met}$ is a homotopy equivalence under $K$. This is clearly true when $K$ is a finite subcomplex of $X$, since then its inclusions $K\subseteq X$ and $K\subseteq X_{Met}$ are cofibrations (the latter following since $K$ is an ANR for metric spaces). That it should remain true for any $K$ (say a Cantor set or compact fractal) is quite surprising!

Answer 2

I had trouble finding a reference online for a non-metrizable CW complex, so I figured it might be nice to record that here. Let $X = \bigvee_{n=1}^\infty S^1$ be the infinite bouquet of circles, considered as a CW complex. I claim that this space is not metrizable. An important fact about CW topology that we can exploit is that any open neighborhood of the basepoint $x_0$ intersects the interior of every 1-cell of this cell complex.

Now suppose $X$ is metrizable, i.e. has a metric $d$ which induces the same topology as that of the cell structure. Any open ball centered at $x_0$ intersects the interior of every 1-cell. So we can pick a sequence $(x_n)_{n=1}^\infty$ such that $x_n$ lies in the interior of the $n$-th 1-cell and $d(x_0,x_n)<1/n$. Then $\lim x_n = x_0$ by construction. On the other hand, the set $X-\{x_1,x_2,\dotsc\}$ is an open neighborhood of $x_0$ since it is open in each cell of the complex. Thus $x_0$ is separated from the sequence $(x_n)_{n=1}^\infty$, a contradiction. $\quad \square$

Answer 3

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).