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

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    $\begingroup$ 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. $\endgroup$ Jan 13, 2013 at 19:49
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    $\begingroup$ 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. $\endgroup$ Jan 14, 2013 at 15:21
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    $\begingroup$ 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. $\endgroup$ Jan 14, 2013 at 21:48
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    $\begingroup$ 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! $\endgroup$ Jan 15, 2013 at 14:30
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    $\begingroup$ Sure. It's Theorem 2.1 in R. Cauty's Rétractions dans les espaces stratifiables. $\endgroup$
    – Tyrone
    Dec 13, 2022 at 13:39

3 Answers 3


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!


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.

  1. 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).

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. $\quad \square$

  • $\begingroup$ That's a nice construction! $\endgroup$
    – David Roberts
    Sep 23, 2021 at 6:24
  • $\begingroup$ Nice. This is exactly the example I note in the comments above. Even though $X$ is not regular you can let $Y=\bigcup_{n\in\mathbb{N}}\{(x,y)\in\mathbb{R}^2\mid (x-1-\frac{1}{n})^2+y^2=(1+\frac{1}{n})^2\}$. Now $Y$ is something of a metrizable version of $X$. There is a continuous bijection $X\to Y$. The homotopy inverse $Y\to X$ quotients by a small closed ball about the origin. $\endgroup$ Sep 23, 2021 at 13:51
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    $\begingroup$ The argument in this answer seems to be using a nonstandard definition of what a cell in a CW complex is. According to the standard definition, as in Whitehead's original paper and most recent sources, a CW complex is the disjoint union of its cells, so cells are "open cells". Unfortunately some authors in the intervening years took "cell" to mean the closure of what Whitehead called a cell. With this definition a "cell" need not be contractible. For example, every smooth closed connected manifold is a "cell" by this definition. $\endgroup$ Sep 23, 2021 at 16:01
  • $\begingroup$ @AllenHatcher are you referring to how I kept saying "the interior of a cell"? I think I just wrote that to avoid any of that sort of confusion. I think the argument works exactly the same regardless of how you interpret "cell", no? $\endgroup$ Sep 23, 2021 at 21:31
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    $\begingroup$ @EthanDlugie: I was referring to your statement that a set is open if it is open in each cell. If this were true with the usual definition of cells then every subcomplex would be an open set. For a set to be open it needs to be open in the closure of each cell. $\endgroup$ Sep 24, 2021 at 23:13

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