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Let's consider a diagram $\Phi: \lambda \to \mathcal{T}_*$ $$ X_0 \to X_1 \to \cdots \to X_\xi \to X_{\xi+1} \to \cdots $$ of pointed spaces, indexed by some ordinal $\lambda$, in which each $X_\xi$ is a subspace of a certain space $Y$ and each map $X_\zeta\to X_\xi$ is the inclusion map.

I have the idea that 'morally' the categorical colimit of the diagram is simply the union $X = \bigcup_{\xi< \lambda} X_\xi$ (with the subspace topology). Certainly this is not actually the case in general -- we need some topological restrictions.

I'm willing to impose really strong topological restrictions. First of all, I'm happy to assume all of the maps $X_\zeta \to X_{\xi}$ and $X_\xi \to Y$ are cofibrations. I'd even be pretty happy to have an argument in which each map $X_\xi \to X_{\xi + 1}$ is obtained by attaching a cell, or a cone. Furthermore, I'm content to work with a category of spaces (such as CGWH) for which cofibrations are necessarily inclusions of closed subspaces (up to homeomorphism).

In general, of course, there is a comparison map $c: \mathrm{colim}\ \Phi\to X$, and it is clearly surjective. It would suffice, therefore, to prove that $c$ is a cofibration.

Unfortunately, my diagram-wrangling skills are coming up short. I'd appreciate pointers to a good argument or a useful reference.

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    $\begingroup$ Isn't that always the case as soon as the union is equiped with the initial topology ? If the problem is that you don't want to use the initial topology, what topology do you put on the union ? $\endgroup$ Commented Jul 29, 2020 at 16:10
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    $\begingroup$ The union gets the subspace topology. $\endgroup$
    – Jeff Strom
    Commented Jul 29, 2020 at 16:15
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    $\begingroup$ how about section 3.2 here: neil-strickland.staff.shef.ac.uk/courses/homotopy/cgwh.pdf ? $\endgroup$ Commented Jul 29, 2020 at 17:03
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    $\begingroup$ Let me know if this doesn't work, I'm too busy right now to think about it. Take $X=\mathbb{N}$ and give it the cofinite topology. Let $X_n=\{1,\dots,n\}$ for each $n\geq1$. Then $X_n$ is discrete, and $X_{n+1}$ is obtained from $X_n$ by attaching a $0$-cell. Each $X_n\subseteq X$ is closed, and I think it it a cofibration (it has a halo in $X$?). As sets $X=\bigcup X_n$, but as spaces $colim\;X_n$ is discrete. Note that everything is compactly generated (although not weak Hausdorff?). $\endgroup$
    – Tyrone
    Commented Jul 29, 2020 at 17:05
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    $\begingroup$ Let $X_n=\{1, 1/2, \ldots, 1/n, 0\}$ as a subspace of $\mathbb{R}$. When we take the colimit, we just get a countable set with the discrete topology, which is different from the subspace topology of the union of the $X_n$'s in $\mathbb{R}$. $\endgroup$ Commented Jul 29, 2020 at 17:11

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