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Consider the diagram of finite CW complexes $X \stackrel{f}\leftarrow Y$ where $f$ is a cellular map and note that its homotopy colimit is precisely the mapping cylinder $$C_H = \frac{X \sqcup (Y \times [0,1])}{\sim}$$ where $(y,0) \sim f(y)$. It is not too difficult to impose a nice CW structure on $C_H$ once you cellulate the unit interval: maybe the hardest thing here is to realize the product of finite CW complexes as a CW complex.

Here's my question:

Is there any hope whatsoever of imposing a (nice) CW structure on the homotopy limit of a dual diagram?

By dual diagram I mean a map $X \stackrel{g}\to Y$ of finite CW complexes where $g$ is any cellular map. Let $Y^I$ be the path space of all continuous $\gamma:[0,1] \to Y$, and note that the homotopy limit of our dual diagram is defined to be $$L_H = \{(x,\gamma) \in X \times Y^I\mid \gamma(0) = g(x)\}.$$ I don't see how to make the path space $Y^I$ a CW complex in any nice way, but maybe we can still do something with the subspace $L_H$.

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Here are a few simple remarks. Regarding the second sentence, you probably need $f$ to be a cellular map in order to get a CW-structure on its mapping cylinder. Otherwise, you probably only get a cell structure. Further, do you allow for non-trivial conditions to be imposed on the function $g$? If not, then taking $g$ to be the identity on $Y$ we get $L_H = Y^I$, which slightly reduces our task. – Ricardo Andrade Jul 24 '13 at 16:24
@RicardoAndrade $f$ and $g$ are to be treated as arbitrary cell maps. It doesn't help too much to treat $g$ as the identity on $Y$: basically, if we could put a CW structure on $Y^I$ itself, then immediately we would get $L_H$ as a subcomplex of $X \times Y^I$ for any cell map $g$. – Vidit Nanda Jul 24 '13 at 16:57
I don't know how to give a rigorous proof, but I believe $L_H$ should almost never be a CW complex--it's infinite-dimensional, in a way that's like a Banach space (essentially, you're looking at a space of functions with the sup metric) rather than like a countably infinite-dimensional vector space that can be topologized as the colimit of its finite dimensional subspaces. However, $L_H$ does always have the homotopy type of a CW complex. Milnor proved a very general result along these lines in his paper "On spaces having the homotopy type of a CW-complex". – Eric Wofsey Jul 24 '13 at 17:37
@EricWofsey yes the path space is awful, but here's a question: do we really need the entire path space to build the homotopy limit, or does it suffice (say) to only consider homotopy classes of paths? – Vidit Nanda Jul 24 '13 at 18:24
Do you really mean to ask whether there is a factorization of $g$ as $X\to Z\to Y$ where the first map is a homotopy equivalence, the second map is a fibration, and $Z$ is a CW complex? If by "fibration" you mean Serre fibration, the answer is yes (by the small object argument construction of the factorization of $g$ into an acyclic cofibration and a fibration). If you want to demand a Hurewicz fibration, the problem seems harder. – Eric Wofsey Jul 24 '13 at 19:07

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