# CW complex structure of geometric realization

In Ralph Cohen's notes on the topology of fiber bundles he makes the following claims:

on pp.69, he says the geometric realization of a simplicial set is a CW complex

on pp.70, he says the geometric realization of a simplicial space may not be a CW complex

My question is, what is in the category of sets that guarantees the geometric realization has a CW complex structure which is missing in the category of topological spaces?

• I'm confused by your confusion. It's sets that are a subcategory of topological spaces; the geometric realization of a simplicial space can be any topological space. – Qiaochu Yuan May 29 '16 at 22:27
• @Qiaochu: you mean the category of sets is a subcategory of the category of topological spaces? – PhysicsMath May 29 '16 at 23:11
• Yes, of course. – Qiaochu Yuan May 29 '16 at 23:14
• Have you read a proof of the fact that the geometric realization of a simplicial set is a CW complex? – Eric Wofsey May 29 '16 at 23:15
• @Physics: a topological space is a set of points with a topology, and the geometric realization of a simplicial space depends on the topology of the spaces that make it up. There is a functor from topological spaces to sets, but it doesn't exhibit topological spaces as a full subcategory, because morphisms of topological spaces are required to be continuous. If it's not clear to you that the second construction is strictly more general than the first then I recommend that you look very closely at the definition of the geometric realization of a simplicial space before going any further. – Qiaochu Yuan May 30 '16 at 4:00

Here's a sketch of the proof why realizations of simplicial sets are CW complexes. Roughly speaking, a CW complex is a space built by one-by-one gluing in new simplices along their boundaries. If $X_\bullet$ is a simplicial set, then $|X_\bullet|$ is built by gluing together the spaces $X_n\times\Delta^n$, where you give $X_n$ the discrete topology. But instead of gluing together these spaces, you can just think of $X_n\times \Delta^n$ as a disjoint union of copies of $\Delta^n$, one for each element of $X_n$, and then glue together these simplices one-by-one. So you are just gluing together a bunch of simplices, and you can check that you get a CW-complex.
(I am of course glossing over some details, the most notable of which is that in a simplicial set (unlike a CW-complex), your simplices might be glued together by degeneracy maps instead of by boundary maps. To avoid this issue, you should let $Y_n\subseteq X_n$ be the set of nondegenerate simplices, and then only glue together simplices corresponding to points of $Y_n$, because all the other ones are degenerate and so don't actually add anything to the geometric realization.)
What goes wrong with this for simplicial spaces? Well, if $X_\bullet$ is a simplicial space, then $|X_\bullet|$ is still built by gluing together the spaces $X_n\times\Delta^n$. But this time $X_n$ may not have the discrete topology, so $X_n\times\Delta^n$ is not just a disjoint union of simplices! So you can't glue in each simplex one-by-one; you have to take the topology of $X_n$ into account, and so it is not at all obvious how you could get a CW-complex structure. In fact, for any space $A$, you can consider the constant simplicial space $X_\bullet$ with $X_n=A$ for all $n$ and every face and degeneracy map the identity, and then $|X_\bullet|$ is just $A$. So every space can be the realization of a simplicial space.