# Mapping of compact set into colimit of space

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!

## 1 Answer

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.