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Given a continuous map $p:E\rightarrow B$ where $B$ is given by a colimit of $B_{0}\subseteq B_{1}\subseteq B_{2}\subseteq\dots$. We get the canonical induced map

$$ colim_{n\in\mathbb{N}} p^{-1}(B_{n})\xrightarrow{v} E$$

It is a continuous and bijective map. The background discussion is: Is it a homeomorphism? What are sufficient conditions for it to be a homeomorphisms/homotopy equivalence/weak homotopy equivalence?

What I can say is, that it is a weak equivalence, if $p$ is a Serre fibration and $E$ is nonempty. For in this case choose an element of $e$ and let $F:=p^{-1}(\{p(b)\})$. The long exact sequence for Serre fibrations yields

$$\require{AMScd} \begin{CD} \dots @>>> \pi_{n}(F, e) @>>> \pi_{n}(colim_{n\in\mathbb{N}}p^{-1}(B_{n}),e) @>>> \pi_{n}(B, p(e)) @>>> \dots \\ & @VV id V @VV \pi_{n}(v) V @VV id V \\ \dots @>>> \pi_{n}(F,e) @>>> \pi_{n}(E,e) @>>> \pi_{n}(B,p(e)) @>>> \dots \\ \end{CD} $$

Now we can apply the five lemma.

My two questions are:

Did I use a sledgehammer to crack a nut? Is there a more elementary way to see this?

Can this statement be generalised or strengthened?

Eric Wofsey
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user520682
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  • Why is $\operatorname{colim} p^{-1}(B_n)\to B$ a Serre fibration? – Eric Wofsey Jul 18 '19 at 20:13
  • I guess that is automatic if $B$ is $T_1$ since then any compact subspace of $B$ is contained in some $B_n$. – Eric Wofsey Jul 18 '19 at 20:18
  • Thanks for pointing out that it is not clear whether $\operatorname{colim}p^{-1}(B_{n})\rightarrow B$ is a Serre fibration. I guess this would be true, if the space B satisfies the assumptions from your answer. But then your proof shows that we can drop the assumption of $p$ being a Serre fibration anyway to show that $v$ is a weak equivalence. – user520682 Jul 19 '19 at 11:41

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As long as every compact subspace of $B$ is contained in some $B_n$ (which is automatically the case if $B$ is $T_1$; see Compact subset in colimit of spaces), then $v$ is a weak equivalence. Indeed, note that this implies every compact subspace $K$ of $E$ is contained in some $p^{-1}(B_n)$, since $p(K)$ is compact. Since homotopy groups are defined in terms of maps out of compact spaces, this immediately implies $v$ is a weak equivalence.

Here is another criterion you may find helpful: if the interiors of the $B_n$ cover $B$, then $v$ is a homeomorphism. Indeed, since $p$ is continuous, this implies the interiors of the $p^{-1}(B_n)$ cover $E$, which implies $v$ is a homeomorphism.

It is not true in general that $v$ is a weak equivalence. For instance, let $B=\mathbb{N}$ with the indiscrete topology and $B_n=\{0,\dots,n-1\}$. Let $E=\mathbb{R}$ and let $p:E\to B$ be a map such that $p^{-1}(\{n\})$ is dense in $\mathbb{R}$ for all $n$. Then $p^{-1}(B_n)$ is totally disconnected for all $n$ (since its complement is dense) and thus the colimit $\operatorname{colim} p^{-1}(B_n)$ is totally path-disconnected (since any path in the colimit would be contained in $p^{-1}(B_n)$ for some $n$ by compactness). In particular, then, $v$ is not an isomorphism on $\pi_0$.

Eric Wofsey
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  • Thanks for the answer, this is what I was looking for! – user520682 Jul 19 '19 at 11:46
  • Yet, I am struggling with the counterexample: You are using that the image of some compact set in $\operatorname{colim}p^{-1}(B_{n})$ lies in one of the $p^{-1}(B_{n})$ (*). But with the argument you gave in the first paragraph of your answer, wouldn't this imply $0\neq\operatorname{colim}\pi_{0}(p^{-1}(B_{n}))\cong\pi_{0}(\operatorname{colim}p^{-1}(B_{n}))\cong\pi_{0}(\mathbb{R})$? Why should (*) hold? – user520682 Jul 19 '19 at 11:53
  • The first paragraph is talking about compact subsets of $E$, not compact subsets of $\operatorname{colim}p^{-1}(B_{n})$. (*) holds just by the result on compact subsets of $T_1$ colimits I referred to at the start. – Eric Wofsey Jul 19 '19 at 14:24
  • I see. I am still wondering why there should exist a decomposition of the reals into countably many disjoint dense subsets. Could you please sketch the idea or give a reference (I couldn't find any)? – user520682 Jul 27 '19 at 09:50
  • That's very easy. Just fix a countable base, and one by one pick a point in each set in the countable base to be in each of your dense subsets. (Then put the leftover points that weren't chosen wherever you want.) In fact, doing this by transfinite induction, you can get a partition into $2^{\aleph_0}$ dense subsets. – Eric Wofsey Jul 27 '19 at 13:29