Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $(G, n)$ be a pair where $G$ is an abelian group and $n \in \mathbb{N}$. Recall that an Eilenberg-MacLane space is a connected CW complex $X$ such that $\pi_r(X) = G$ if $r=n$ and $0$ otherwise. An Eilenberg-MacLane space is unique up to homotopy equivalence. A direct argument is given in Hatcher's book based on the Whitehead theorem (that a weak equivalence of CW complexes is a homotopy equivalence): one shows that there is always an Eilenberg-MacLane space of a given form (that starts with a wedge of $r$-spheres and then goes up). Then, one explicitly checks that given any Eilenberg-MacLane space, one can define a map from this one into the other one which induces an isomorphism on the homotopy group in dimension $n$, though. This relies on a particular construction of the Eilenberg-MacLane space, though.

Is there a more functorial way of seeing this?

One half of an approach is the following. We know that the Eilenberg-MacLane spaces represent the cohomology functors (with coefficients in $G$) on $CW_*$, the homotopy category of pointed CW complexes. This implies uniqueness.

But I don't think this is logically correct given the above constraint. Namely, the fact that any Eilenberg-MacLane space represents the cohomology functor relies, I think, on the fact that any two are homotopy equivalent. Namely, the proof I know of this fits the various $\{ K(G,n) \}$ into an $\Omega$-spectrum and then argues that they represent a cohomology theory which satisfies the dimension axiom, so is singular cohomology with coefficients. I don't see how it is obvious that any $K(G,n)$ can be fit into an $\Omega$-spectrum, though, without using the homotopy equivalence above.

share|improve this question
What do you mean by functorial proof? How about working in simplicial sets, where there are canonical and functorial constructions and then using the Quillen equivalence to Top? –  David Roberts Dec 9 '10 at 22:33
@David: One that does not use the explicit expression of an Eilenberg-MacLane space. A proof via simplicial sets would be fine, but I'd be curious about a proof that the explicit functorial construction is the unique one (up to homotopy equivalence, for fibrant simplicial sets, say). –  Akhil Mathew Dec 9 '10 at 23:18
I would that the argument in Hatcher does not depend on a specific construction of $K(A,n)$. Given any $K(A,n)$ $X$ that is a CW-complex, the inclusion of the $n-1$-skeleton $X^{(n-1)}$ into $X$ is nullhomotopic by obstruction theory, so we can replace $X$ by $X/X^{(n-1)}$ and then apply Hatcher's argument. This gives a homotopy equivalence between any two CW-complexes that are $K(A,n)$s, without passing through any particular special favorite $K(A,n)$. –  Eric Wofsey Dec 9 '10 at 23:36
This is obvious in the case of simplicial sets, since the homotopy of a simplicial abelian group is isomorphic to its homology). –  Harry Gindi Dec 10 '10 at 6:40
add comment

3 Answers

up vote 11 down vote accepted

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.

  1. There are no obstructions to defining the map on the 0-skeleton (send everything to the basepoint).
  2. The map from $\pi_1(X)$ to $\pi_1(Y)$ tells you how to define the map on the 1-skeleton.
  3. The fact that the map is a homomorphism means you can extend it to the 2-skeleton.
  4. 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.

  5. (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.

share|improve this answer
Thanks! . –  Akhil Mathew Dec 11 '10 at 5:11
Hi, Tom! Your answer confused me a bit. I think it's important to point out that for $n>1$, the correct version of Proposition 1B.9 is that for any (n-1)-connected CW complex X, the maps on $\pi_n$ are indueced by maps $X \to K(G,n)$. In this case, the maps on $\pi_n$ are basically the same as $n$-th cohomology of $X$. If you don't say "$(n-1)$-connected", the statement is wrong. For example, take $n=3$ and $X=S^2$. –  Ilya Grigoriev Oct 14 '11 at 6:08
Dear Ilya, you're completely correct. That's a good thing to point out explicitly. Thanks! –  Tom Church Oct 14 '11 at 15:31
add comment

You can construct $K(G,n)$ functorially as the geometric realization of the simplicial set whose set of $k$-simplices is the collection of closed $G$-valued degree $n$ cochains on $\Delta^k$.

Example: $K(G,1)$.
{ closed $G$-valued $1$-cochains on $\Delta^0$ } = $ \{ * \} $.
{ closed $G$-valued $1$-cochains on $\Delta^1$ } = $G$.
{ closed $G$-valued $1$-cochains on $\Delta^2$ } = $G^2$.

Example: $K(A,2)$.
{ closed $A$-valued $2$-cochains on $\Delta^0$ } = $ \{ * \} $.
{ closed $A$-valued $2$-cochains on $\Delta^1$ } = $ \{ * \}$.
{ closed $A$-valued $2$-cochains on $\Delta^2$ } = $A$.
{ closed $A$-valued $2$-cochains on $\Delta^3$ } = $A^3$.

share|improve this answer
Does this help show why any other model of a K(A,n) is homotopic to the canonical one? I imagine it might, but I'm being a bit thick at the moment (see also my comment on the question) –  David Roberts Dec 9 '10 at 22:34
It is NOT true that any other model is canonically homotopy equivalent to this one. Indeed, unless $n=1$, the connected components of the space of (pointed) maps between two homotopy equivalent Eilenberg-MacLane spaces $K(A,n)$ is not contractible. If you work at the level of the homotopy category, i.e., identify homotopic maps, then, of course, the problem goes away. –  André Henriques Dec 10 '10 at 19:47
Sorry. Scratch what I said above. The homotopy groups of $Map(H(A,n),H(A,n))$ are all in negative degrees. So the space of (pointed) maps between two homotopy equivalent Eilenberg-MacLane spaces $K(A,n)$ is actually contractible. –  André Henriques Dec 10 '10 at 19:55
Out of curiosity, is it any easier to prove the corresponding result for simplicial sets? –  Akhil Mathew Dec 11 '10 at 5:12
(Harry Gindi's comment to the question takes care of the case of simplicial abelian groups, if I understand correctly.) –  Akhil Mathew Dec 11 '10 at 5:15
add comment

If you accept the Whitehead theorem, then simple obstruction theory (without cohomology) determines the homotopy type of spaces with homotopy groups in only one dimension. That is, you can easily show that if the only nonzero homotopy group of the CW complex $Y$ is in dimension $n$, then for any $(n-1)$-connected CW complex $$ [X, Y] \cong \mathrm{Hom}( \pi_n(X), \pi_n(Y) ) . $$ Then if the homotopy groups of $X$ are also concentrated in dimension $n$, an isomorphism $\pi_n(X) \xrightarrow{\cong} \pi_n(Y)$ gives rise to a homotopy equivalence $X\xrightarrow{\simeq} Y$.

But what if we don't want to use Whitehead? I think you are stuck, because basic (model category theoretic) arguments can be used to prove Whitehead (at least for CW complexes with finitely many finitely generated homotopy groups?) from the uniqueness of Eilenberg-Mac Lane spaces.

(Prove it by induction for CW complexes with at most $n$ nonzero homotopy groups; this requires us to convert maps to fibrations, and we don't want to end up with non-CW complexes, which is why I said we might need finitely generated homotopy groups.)

CLARIFICATION: When I said "model-theoretic" I meant the basic facts about cofibrations, fibrations, induced maps, etc. The finite generation comes in when we convert a map to a fibration: this is done with a path space and a pullback, and these things can easily take us away from CW complexes, but my plan was to impose conditions that would guarantee we can apply Milnor's theorem (on loops of CW complexes and so on) to the CW complexes in question.

share|improve this answer
Sorry, I'm a bit confused. The model-categorical version of Whitehead that I know is that a weak equivalence in a closed model category between cofibrant-fibrant objects is a homotopy equivalence, and is proved by a bit of diagram-chasing (I think the lemma of Ken Brown does it as well). In the present case, one just uses the fact that any CW complex is cofibrant and fibrant --- I'm not sure how this requires finite generation of the homotopy groups or Eilenberg-MacLane spaces. –  Akhil Mathew Dec 11 '10 at 5:22
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.