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I know that if $X$ is a topological space such that $X= \underset{i}{\bigcup} X_i$ where $X_0 \subset X_1 \subset ... \subset X_n \subset ...$, where $X_i$ are all hausdorff, then the functor $\pi_n(\_)$ commutes with the colimit: $$\varinjlim \pi_n(X_i, x_0)\cong \pi_n(X, x_0) $$ One way to prove this is to show that each continuous map from a compact space $K$ into $X$ factors over some $X_i$.

Can this be generalized to all filtered colimits? Thus, if $I$ is a filtered category, F a functor $I \longrightarrow$ Top, does it generally hold that: $$\varinjlim \pi_n(F(i), x_0)\cong \pi_n(\varinjlim F, x_0) $$ If yes, how can I prove this? I have tried to generalize the proof of the above statement, but I do not know how to prove that each continuous map from a compact space K into the limit factors over some $F(i)$.

stoneaa
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Balletti
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  • "functor $_$ commutes with the colimit"

    Please tell me in which textbook I can find this fact

     – Arshak Aivazian Jan 16 '21 at 11:20
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    Years later: I feel like there is something subtle about the initial statement. If $X=\mathbb R/\mathbb Z$ and $X_n$ is the image of $[0,1-2^{-n}]$, then $\pi_1(X)\cong\mathbb Z$ is not the inductive limit of the $\pi_1(X_n)\cong0$. What I assume the statement says is that the union has to realize the inductive limit of the injections between the $X_n$, in the category of topological spaces, which is much more of a subtle concept than mere set-theoretic union. – Pierre PC Jun 06 '25 at 10:39

1 Answers1

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The answer is no, not even $\pi_1$ does:

If $X$ is a metric space and $Y$ any topological space, then a map $f\colon X\to Y$ is continuous if and only if for every convergent sequence $(x_n)_{n\in\mathbb{N}}$ in $X$, the sequence $(f(x_n))_{n\in\mathbb{N}}$ converges with limit point $f(\lim_n x_n)$. Furthermore, if $X$ is compact, the closure of the set underlying a convergent sequence is closed and countable. Thus, such an $X$ is naturally isomorphic to the filtered colimit of all of its countable, closed subsets. Fixing a point $x_0\in X$, we may equivalently restrict the indexing category to the set of countable, closed subsets containing $x_0$.

Now take a space as easy as $X = S^1$ with any point $x_0\in X$. We have $\pi_1(A,x_0) = 1$ for each closed and countable subset $A\subset X$, but $\pi_1(X,x_0) = \mathbb{Z}$, of course.

Ben
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  • Thanks, that is very helpful. Would the statement hold if one restricted X to a CW complex and all F(i) to be subcomplexes? – Balletti May 18 '17 at 07:24
  • For filtered limits by sub-complexes, this stands a chance to hold, but I'm not entirely sure. Have you checked May's Concise Course? – Ben May 18 '17 at 09:04
  • Dear @Balletti, in fact, if $(X,x_0)$ is CW, then (based) maps from spheres (and disks, respectively) into $X$ can up to homotopy be replaced by cellular maps. Furthermore, if $X$ is a filtered colimit of sub-complexes, every cellular map (or null-homotopy, resp.) into $X$ is detected on some element of the filtration. – Ben May 18 '17 at 11:57
  • Thanks, how would I show that every cellular map is detected by some element of the filtration? Do you have a reference for this? – Balletti May 19 '17 at 13:26
  • Oh, I'm afraid I was assuming some kind of finiteness, which probably makes it possible to reduced every filtered colimit of CW-complexes to a plain filtration. I'm not so sure any more, sorry. – Ben May 19 '17 at 13:38
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    @Balletti A compact set only meets finitely many cells in a CW complex. Is this the finiteness condition needed? – Trebor Dec 08 '23 at 12:13