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Jul 24, 2013 at 19:07 comment added Eric Wofsey 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.
Jul 24, 2013 at 18:24 comment added Vidit Nanda @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?
Jul 24, 2013 at 17:37 comment added Eric Wofsey 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".
Jul 24, 2013 at 16:57 comment added Vidit Nanda @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 $g$.
Jul 24, 2013 at 16:53 history edited Vidit Nanda CC BY-SA 3.0
edited to account for ricardo's comments
Jul 24, 2013 at 16:24 comment added Ricardo Andrade 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.
Jul 24, 2013 at 13:15 history asked Vidit Nanda CC BY-SA 3.0