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I am looking for a reference or a proof of the following fact:

Let $X_{1}\subset X_{2}\subset\dots $ be a sequence of (hausdorff) topological spaces indexed by natural numbers such that each $X_{i}\subset X_{i+1}$ is a closed subspace for any $i\in \mathbb{N}$. We define $X=colim_{i\in \mathbb{N}}X_{i}$.

Then the $H_{m}(X,\mathbb{Z})=colim_{i\in \mathbb{N}}H_{m}(X_{i},\mathbb{Z})$ for any natural number $m\in\mathbb{N}$.

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This is a Theorem on page 115 of Peter May's book A concise course in algebraic topology.

A discussion can be found here.

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  • $\begingroup$ @Thanks for the references, as far as I understand, your first reference does not answer explicitly my question. The discussion you linked, says something interesting. the condition for which it becomes true is : if any map $K\rightarrow X$ (K compact) factors through some $X_{i}$. Is it true in the case of my question ? $\endgroup$
    – Let
    Commented May 29, 2019 at 17:28
  • $\begingroup$ The reference does in fact answer your question. And yes, it is true $\endgroup$
    – Maxime Ramzi
    Commented May 29, 2019 at 17:37
  • $\begingroup$ @Max in the reference it is written "union of an expanding sequence of subspaces" what does it mean exactly ? $\endgroup$
    – Let
    Commented May 29, 2019 at 17:45
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    $\begingroup$ @Max I do believe that the condition "if any map $K\rightarrow X$ ($K$ compact) factors through some $X_{i}$" is essential, it is not always verified. I do believe that compact spaces are small with respect to closed inclusion of Hausdorff spaces but not small for arbitrary inclusions. $\endgroup$
    – Let
    Commented May 29, 2019 at 17:53
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    $\begingroup$ It is possible that May has some hidden assumptions that imply your condition, for instance if he is working with compactly generated spaces. He states : "From here on, we agree that all given spaces are to be compactly generated, and we agree to redefine any construction on spaces by applying the functor k to it". With this condition, you get the thing about compact spaces : the proof given here (math.stackexchange.com/questions/1584667/…) indeed works just as well $\endgroup$
    – Maxime Ramzi
    Commented May 29, 2019 at 18:08
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Here is a reference, proposition 2.4.2 page 49.

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