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YCor
YCor
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It$\newcommand\met{\mathrm{met}}$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}$$X_{\met}$ (with the same underlying set), so that the continuous identity function $X\to X_{met}$$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.

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.

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

Although I didn't forget about this question, I did neglect to make my question more detailed and explicit after the discussion in the comments. This has finally been done.
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Jeremy Brazas
Jeremy Brazas
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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 finite dimensional CW-complex (possibly with countably many cells in eachand maybe even of finite dimension). Is it possible weaken the topology of $X$ to construct another space $X_{met}$ (with the same underlying set), without changingso that the continuous identity function $X\to X_{met}$ is a homotopy typeequivalence?

Update: I will clarify (now much later) that this question has an affirmative answer for simplicial complexes. Given an arbitrary simplicial complex $K$, sowe have $|K|$, which has the resulting spaceweak 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.

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?

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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Jeremy Brazas
Jeremy Brazas
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Making CW-complexes metrizable

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?