[Math] CW-complex of Eilenberg-MacLane spaces

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What is the CW-complex of Eilenberg-MacLane space $K(\mathbb{Z}_2,2)$?

What is the CW-complex of Eilenberg-MacLane space $K(\mathbb{Z}_n,d)$?

What is the CW-complex of Eilenberg-MacLane space $K(\mathbb{Z}_n\times \mathbb{Z}_m,d)$?

For example, I like to know the number of cells in each dimensions, and the chain complex formed by those cells, so that one can compute cohomology.
(I like to know the chain complex, in addition to the cohomology).

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There is a completely explicit simplicial set realizing to $K(A,n)$ for each $n$, coming from the Dold-Kan correspondence.

It is defined as $$\bar{A}[S^n]=\mathrm{ker}(A[S^n]\to A[*])$$ (the kernel is taken levelwise, and $A[X]$ for each simplicial set $X$ is obtained by taking the free $A$-module levelwise). Here I let $S^n=\Delta^n/\partial \Delta^n$.

I don't think you can get more explicit than that.

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[Math] “Dirty” proof that Eilenberg-MacLane spaces represent cohomology

I'd suggest looking up some basic material on obstruction theory. There, you generally find classification of maps $X \to Y$ with domain a CW-complex in terms of cohomology groups $H^s(X;\pi_t(Y))$. The proofs are often very cellular indeed.

In the case where the range is an Eilenberg-Maclane space (for an abelian group), the dirty proof is something like:

Any map from $X$ is homotopic to one where the (n-1)-skeleton $X^{(n-1)}$ maps to the basepoint of $Y$.
A map on the n-skeleton $X^{(n)}$ sending the (n-1)-skeleton to the basepoint is determined, up to homotopy, by a choice of element of $G$ for each n-cell of $X$, essentially by definition of homotopy. This is an element in the n'th CW-chain group $C^n_{CW}(X;G)$.
Such a map extends to all higher skeleta if and only if the attaching maps for all the (n+1)-cells become nullhomotopic in $Y$. Thus the map extends if and only if it's represented by a cocycle, i.e. an element of $Z^n_{CW}(X;G)$.
This is a complete invariant, up to homotopy, of maps that are trivial on the (n-1)-skeleton. (Higher cells have basically unique maps up to homotopy.)
Any homotopy between two such maps can be pushed to a homotopy that's trivial on the (n-2)-skeleton of $X$.
Such a homotopy is determined, up to a "track" (a homotopy between homotopies), by a choice of element of $G$ for each (n-1)-cell of $X$.
Such a homotopy alters the map on the n-skeleton (as an element of $C^n_{CW}(X;G)$) by adding a coboundary element, something in $B^n_{CW}(X;G)$.
Therefore, the full mapping space mod homotopy is $H^n_{CW}(X;G)$.

This is a little messy. Often it's nice to use the filtration of $X$ by subcomplexes $X^{(n)}$ and use that each inclusion in the filtration induces a fibration of mapping spaces $$F(X^{(n)}/X^{(n-1)},Y) \to F(X^{(n)},Y) \to F(X^{(n-1)},Y)$$ to clean this homotopical analysis up a little into something slightly more systematic. This leads to a spectral sequence for the homotopy groups of the mapping spaces in terms of the cohomology of $X$ with coefficients in the homotopy groups of $Y$, but you have to be a little careful because there is a "fringe" that exhibits some non-abelian-group-like behavior.

Eilenberg-MacLane Spaces – Functorial Proof of Uniqueness Up to Homotopy

I think you're referring to the argument on page 366 of Hatcher. But there is an earlier argument for $n=1$ in Chapter 1B, which works just as well in general (as Eric Wofsey points out above).

Hatcher deduces

Theorem 1B.8: The homotopy type of a CW complex $K(G,1)$ is uniquely determined by $G$.

from the following (which is the "functorial" perspective on Eilenberg-MacLane spaces):

Proposition 1B.9. Let $X$ be a connected CW complex and let $Y$ be a $K(G,1)$. Then every homomorphism $\pi_1(X,x_0)\to \pi_1(Y,y_0)$ is induced by a map $(X,x_0)\to (Y,y_0)$ that is unique up to homotopy fixing $x_0$.

I won't spell out the details of the argument (you should just read pages 90–91 of Hatcher), but roughly, the proof of the proposition breaks up into the following steps.

There are no obstructions to defining the map on the 0-skeleton (send everything to the basepoint).
The map from $\pi_1(X)$ to $\pi_1(Y)$ tells you how to define the map on the 1-skeleton.
The fact that the map is a homomorphism means you can extend it to the 2-skeleton.
There are no obstructions to extending the map to the 3-skeleton, the 4-skeleton, etc., since any map of the boundary of a ≥3-cell (i.e. a ≥2-sphere) into $Y$ extends over the whole cell.
(For uniqueness) Any two such maps are homotopic on the 1-skeleton, since the map on $\pi_1(X)$ is specified. This lets you extend any map of $X\times$$\{$$0,1$$\}$ to the 1-skeleton of $X\times [0,1]$. To extend to all of $X\times [0,1]$ and thus prove the maps are homotopic, apply Steps 3 and 4 again.

Once you understand this argument, I think it is definitely worth understanding for yourself why the same steps work for a general $K(G,n)$. The lemma you mention (Lemma 4.31, page 366) is exactly the higher-dimensional analogue of Step 3.

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[Math] “Dirty” proof that Eilenberg-MacLane spaces represent cohomology

Eilenberg-MacLane Spaces – Functorial Proof of Uniqueness Up to Homotopy

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