-1
$\begingroup$

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 $m\in\mathbb{N}$.

| cite | improve this question | | | | |
$\endgroup$
2
$\begingroup$

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

A discussion can be found here.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ @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 ? $\endgroup$ – TTip May 29 '19 at 17:28
  • $\begingroup$ The reference does in fact answer your question. And yes, it is true $\endgroup$ – Maxime Ramzi May 29 '19 at 17:37
  • $\begingroup$ @Max in the reference it is written "union of an expanding sequence of subspaces" what does it mean exactly ? $\endgroup$ – TTip May 29 '19 at 17:45
  • 1
    $\begingroup$ @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. $\endgroup$ – TTip May 29 '19 at 17:53
  • 2
    $\begingroup$ 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 May 29 '19 at 18:08
1
$\begingroup$

Here is a reference, proposition 2.4.2 page 49.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.