# Eilenberg-MacLane Spaces - uniqueness?

Given a positive integer $n$ and a group $G$ (necessarily abelian if $n > 2$) then a based space $X$ is said to be $K(G, n)$ if $\pi_k(X)$ is trivial for $k \neq n$ and $\pi_n(X) = G$. First a pedantic question - is such a space necessarily a CW complex (as the wiki seems to suggest) or do we include that in the definition?

Secondly, for fixed $n$ and $G$, why is such a space unique up to homotopy equivalence? I know Whitehead's theorem that says that a map between two CW complexes that induces isomorphisms on all of the homotopy groups (and a bijection on $\pi_0$) is necessarily a homotopy equivalence. So if we have two (CW complex) $K(G,n)$'s say $X$ and $Y$ and we can cook up a map $f: X \to Y$ with $f_\# : \pi_n(X) \to \pi_n(Y)$ an isomorphism, then by Whitehead $X$ and $Y$ are homotopy equivalent. However, why does such a map $f$ exist?

If $H$ is an infinite-dimensional Hilbert space, then $H$ is contractible, so it is a $K(G,1)$ with $G$ the trivial group, but not a CW-complex in any way.

• I fail to see how one cannot say that H has the structure of a CW complex, is there some sense in which adding $D^\infty$ as an $\infty$-dim cell breaks the usual definition of a CW-complex, it seems like it would be consistent with all of the other definitions and usual properties (I'm also not really convinced that you can't construct this with CW-complexes, for instance: $\mathbb{RP}^\infty$ can't be a subset of a finite dimensional space but has a cohomology (and thus one can find its cohomology and write a cw-complex that it's homotopy equivalent to.) – Benjamin Gadoua Oct 14 '16 at 5:21
• That infinite dimensional cell you talk about makes no sense, I am afraid. The definition of CW complexes allows for no such thing. – Mariano Suárez-Álvarez Oct 14 '16 at 5:36
• I suggest you google a bit if "you are not really convinced". It is a somewhat standard fact that an infinite dimensional Hilbert space is not a CW complex —the obstruction is in that it is a Baire space. – Mariano Suárez-Álvarez Oct 14 '16 at 5:36
• I understand that it does not fall into the usual definition where $e^{n}_\alpha$ is a $n$-cell where $n$ is an integer, $\alpha$ lives in some indexing set. What I'm not sure about is how this fails to extend the definition in a consistent manner. – Benjamin Gadoua Oct 14 '16 at 5:38
• You can extend the definition as much as you want, but the definition of CW complexes allows for no such thing. If you intend to use the term "CW complex" in the sense of "whatever definition one might come up with which generalizes real CW spaces" then I doubt anything sensible can be said about your question. – Mariano Suárez-Álvarez Oct 14 '16 at 5:39

For G a nontrivial abelian group we can construct $K(G,n)$ as a wedge of $n$-spheres along with $(n+1)$-cells corresponding to the relations of $G$, and furthermore also adding appropriate $k$-cells for $k>n$ to "close off" all higher-dimensional loops (this last bit is to ensure that $\pi_k(K(G,n))=0$ for every $k>n$). This is a CW complex, and given any other CW complex being a "$K(G,n)$", we can explicitly construct your desired function $f$ inducing isomorphisms on all homotopy groups. See the details in this note: http://www.people.fas.harvard.edu/~xiyin/Site/Notes_files/AT.pdf

EDIT: Added the extra bit about killing off higher loops, as pointed out by the comment by JHF below.

• This gives a Moore space, not an Eilenberg-Mac Lane space. You'll need to attach more cells to kill off the higher homotopy as well. – JHF Oct 13 '16 at 22:56
• Oh right, my bad! Yes, we attach higher-dimensional cells to "close off" all (higher-dimensional) loops as well, yielding trivial $\pi_k$ groups for every $k>n$. Thanks @JHF. – Dan Saattrup Nielsen Oct 14 '16 at 10:52

If you have two $$K(G,n)$$'s, $$X$$ and $$Y$$, then note that $$H_n(X;G)= G$$ by Hurewicz, and then $$H^n(X;G)=G$$ by universal coefficients. Then any $$K(G,n)$$ classifies cohomology so $$\hom(X,Y) =G$$. Then the identity element of $$G$$ corresponds to a natural map between $$X$$ and $$Y$$. Repeating the argument with $$X$$ and $$Y$$ switched shows that this map has an inverse up to homotopy.