3
$\begingroup$

I am reading proposition 2.5.4 of Peter May's 'More Concise Algebraic Topology' (http://www.maths.ed.ac.uk/~aar/papers/mayponto.pdf), a part of which says that

Let $\mathcal{D}$ be a filtered category, and $X_*:\mathcal{D} \to \mathcal{U}$ be a diagram of closed inclusions inside the category $\mathcal{U}$ of compactly generated weak Hausdorff spaces. Let $X$ be the colimit of $X_*$ then for every map $f:K \to X$ of a compact space $K$ into $X$, the image of $K$ lies inside some $X_d$.

May proves this by first constructing $d_n \in \mathcal{D}, k_n \in K, (n=0,1,...)$ such that there are arrows $d_{n-1} \to d_n$ and $f(k_n) \in X_{d_n} \setminus X_{d_{n-1}}$. Then he claims that the countable ordered set $\{ d_n \}$, considered as a subcategory of $\mathcal{D}$, is cofinal in $\mathcal{D}$, hence $X=\varinjlim X_{d_n}$.

This is where I got confused. A countably infinite totally ordered subcategory in a filtered category may not necessarily be cofinal. For example, $\{ \{1 \}, \{1,2 \}, \{1,2,3 \},... \}$ in the directed system of finite subsets of $\mathbb{Z}$. So May's claim must have used some of the topological information, but which I cannot see how.

A similar question has been asked in this post: Compact subset in colimit of spaces, but it is the case when $\mathcal{D}$ is the poset of natural numbers, where a countably infinite subset is indeed cofinal. The answer in that question also doesn't generalize to prove the proposition in May's book.

Any help is appreciated!

$\endgroup$
2
$\begingroup$

The statement of Proposition 2.5.4 is incorrect. Indeed, a counterexample to it is given in Example 2.5.5 immediately following! The space $X=[0,1]$ is the colimit of its countable closed subsets, but $X$ itself is compact and is not contained in any of them.

To correct the hypotheses and get a correct statement, you should assume that that $\mathcal{D}$ is filtered and either:

  • $\mathcal{D}$ has a countable cofinal subset (in which case you can arrange that the sequence $(d_n)$ constructed in the proof is also cofinal), or
  • $\mathcal{D}$ has sequential colimits and $X_*$ preserves them.
$\endgroup$

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.