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Suppose that we have a diagram in the category Top of the following form $$ \cdots X_2 \to X_1 \to X_0=X $$ where the arrows are homeomorphisms. Is it true that that the natural morphism $lim_i X_i \to X$ is a homeomorphism?

It certainly seems like it should be to me, but my intuition sometimes isn't great in these things.

The reason why I am worrying about this is that I wish to show that the infinite barycentric subdivision of a CW complex is homeomorphic to the original complex. Thanks a million in advance for your help!

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Yes. The inverse limit $X' = \lim\{X_i, p_i : X_{i+1} \to X_i\}$ comes along with a family of maps $p'_i : X' \to X_i$ such that the universal property of the inverse limit is satisfied (see e.g. https://en.wikipedia.org/wiki/Inverse_limit).

In your case take $X' = X = X_0$ and $p'_i = p_{i-1}^{-1} \circ \ldots \circ p_0^{-1}$ and verify that the universal property is satisfied.

In your question you also ask for the "infinite barycentric subdivision of a CW-complex". I have never heard about the barycentric subdivision of a CW-complex. Usually this only applies to simplicial complexes. But even in that case I have no idea what an "infinite barycentric subdivision" should be. You can form the $n$-th barycentric subdivision for each $n$, but you cannot get something like an $\infty$-th barycentric subdivision. All simplices of such a space would have mesh $0$, i.e. would be points.

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  • $\begingroup$ Surely you can take the sequential limit of ... -->sd^n(X)-> ...sd(X) -> X? $\endgroup$ – Mathstudent1996 May 16 at 3:21
  • $\begingroup$ Yes, you can do that. But as a topological space the result is $X$ itself and it does not have the structure of a "simplicial complex with infinitely small simplices". $\endgroup$ – Paul Frost May 16 at 8:34
  • $\begingroup$ I wanted to carry out this construction in the category of simplicial sets. I was worried that the realization of sd^oo(X) would be the completion of the realization of X and thus have the wrong homotopy type, but I think that your response shows that this is not the case. $\endgroup$ – Mathstudent1996 May 16 at 9:18
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Yes.

Simply show that $X$ satisfies the universal property of being the limit.

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