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You can build them using a homology decomposition and read off the number of cells in each dimension from the homology groups. For any simply-connected space, this will give you a construction by it …

I think isos up to dimension $n$ is enough to deduce $f_*:[A,X]\to[A,Y]$ onto when $dim(A)\leq n$. This gives a right inverse to $f$ which also induces isos up to $n$ and hence has a right inverse. S …

Let $X$ and $Y$ be CW complexes. Inside of the space of maps $\mathrm{map}(X,Y)$, we have the subspace $\mathrm{CW}(X,Y)$, consisting of just the cellular maps from $X$ to $Y$. The Cellular Approxim …

I don't have a reference, but here is an easier argument, based, like John's, on the homotopy invariance of the homotopy pushout. The invariance implies that you can replace the maps $i: A\to B$ and …

If $X$ is simply-connected, then the homotopy groups will be countable iff the homology groups are countable; and then one can build $X$ by a homology resolution using Moore spaces for countable group …

Map the function $Y\to *$ into the left edge to get two homotopy pushouts in a row; this would produce a cofiber sequence $Y\to X_n\to X_{n+r}$. But this can't be done in general. One interesting w …

In a recent paper I wrote with John Oprea (Oprea, John; Strom, Jeff Lusternik-Schnirelmann category, complements of skeleta and a theorem of Dranishnikov. Algebr. Geom. Topol. 10 (2010), no. 2, 1165–1 …

I think the topological case is handled by this: Brown, Morton A mapping theorem for untriangulated manifolds. 1962 Topology of 3-manifolds and related topics (Proc. The Univ. of Georgia Institute, 1 …