So, today I’m going to talk about Eilenberg-MacLane spaces. These are extremely important in algebraic topology for a number of reasons. The first is that, homotopically, they are simple. By definition, they have only one non-zero homotopy group. So they are in some sense simple.

Moreover, the construction of Postnikov towers shows that any space can be thought of as—in some sense—a “twisted product” of these Eilenberg-MacLane spaces, and in that sense they are “building blocks.” By using the Serre spectral sequence for a fibration, one can often prove things for general spaces by proving them for Eilenberg-MacLane spaces. An important example of this phenomenon is the theorem of Serre (a part of his “mod {C} theory”) that a simply connected space has finitely generated homotopy groups if and only if it has finitely generated homology groups.

Another reason is that many functors in algebraic topology are representable on the homotopy category, at least for CW complexes. By definition, the homotopy groups are representable as maps (in the homotopy category) out of spheres into a space. But more generally, this is true for singular cohomology, and in fact for generalized cohomology theories. The precise statement of this phenomenon is called the Brown representability theorem. I plan to blog more about this more, but a consequence of this is that singular cohomology {H^n(-,  G)} with coefficients in some abelian group {G} is representable. The representing objects are in fact the Eilenberg-Maclane spaces {K(G, n)}.

For now, let’s just assume that the {n}-th cohomology functor {X \rightarrow H^n(X,  G)} is representable on the homotopy category of pointed CW complexes. This means that there is a space {Q(G,  n)} such that there is a natural bijection

\displaystyle  \left\{\mathrm{homotopy \ classes \ } X \rightarrow Q(G,n)\right\} \simeq H^n(X, G).

In particular, by general nonsense, there is a “universal” cohomology class {q_n \in H^n(Q(G, n), G)} such that every cohomology class in {X} is obtained by pulling this back in a manner unique up to homotopy.

Let’s try to determine what the homotopy groups of {Q(G,n)} are. In fact, {\pi_k(Q(G,n))} is the group of homotopy classes of maps {S^k \rightarrow Q(G,n)}. This is in bijection with {H^n(S^k, G)}. And this is zero unless {k=n}, in which it is—as a set—{G}. One can check (e.g. by the Eckmann-Hilton argument) that the group structures are in fact compatbile, though this is irrelevant for the present heuristic discussion. The point is that the homotopy groups of {Q(G,n)} are very simple. Namely,

\displaystyle  \pi_k(Q(G,n)) = G \ \mathrm{if \ } k = n,\quad \ 0 \ \mathrm{otherwise}.

Motivated by this, we make the following definition (and switch to standard notation):

Definition 1 An Eilenberg-MacLane space {K(G,n)} is a connected CW complex such that {\pi_k(K(G,n))} is {G} in dimension {n} and zero otherwise.


We have seen that if something represents the cohomology functors, then it is an Eilenberg-MacLane space. Since we have not proved the Brown representability theorem yet, it will be instructive to take a less fancy approach to constructing these spaces. In fact, we shall eventually show that they are unique.



But first, we should perhaps do a few examples. A good way of constructing {K(G,1)}‘s is to take any contractible space and quotient by a discrete action of {G}. More precisely, any space whose universal cover is contractible is a {K(G,1)} (for {G  =\pi_1}). One trivial example is {S^1}, which is a {K(\mathbb{Z},1)}.

As a more interesting example, the infinite sphere {S^{\infty}} is contractible. This follows from the Whitehead theorem since all the homotopy groups of this vanish. The group {\mathbb{Z}/2} acts on {S^{\infty}} by the antipodal map. The quotient space is {\mathbb{RP}^{\infty}}, and it is thus a {K(\mathbb{Z}/2, 1)}.

More generally, the cyclic group {\mathbb{Z}/m} acts on {S^{\infty}} when thought of as a direct limit of spheres in complex vector spaces {\mathbb{C}^n}, where the action is by multiplication of each coordinate by {e^{2\pi i /m}}. The quotient is an infinite-dimensional lens space, which is a {K(\mathbb{Z}/m, 1)}.

Now it is a fairly straightforward observation that:

Lemma 2 The product of a {K(G,  n)} and a {K(G', n)} is a {K(G \times G', n)}.

The proof is very easy. Namely, the homotopy groups commute with products—because they are representable functors and consequently commute with finite limits in the (pointed) homotopy category. Usual products are products in the homotopy category.

So we have constructed a {K(\mathbb{Z}, 1)} and a {K(\mathbb{Z}/m, 1)}. Thus we can construct a {K(G, 1)} for any finitely generated {G}. Note that if {G} has torsion, then the {K(G, 1)} contains a lens space in the product, and lens spaces have nontrivial cohomology in infinitely many dimensions. Since, as we will see, a {K(G, 1)} is unique up to homotopy equivalence, a finite-dimensional {K(G,  1)} cannot exist if {G} is finitely generated has torsion.

Finally, we should make a brief observation about how the suspension and loop space functors {\Omega} and {\Sigma} interact with Eilenberg-MacLane spaces. Namely, recall that if {A,B} are pointed spaces, then there is natural bijection

\displaystyle  \hom_{\mathbf{PT}}(\Sigma A, B) \simeq \hom_{\mathbf{PT}}(A, \Omega B)

where {\mathbf{PT}} is the homotopy category of pointed topological spaces. We also know that {\Sigma S^n =  S^{n+1}}.

In particular:

Proposition 3 If {X} is a {K(G, n)}, then {\Omega  X} has the homotopy type of a {K(G,  n-1)}.


Indeed, it is clear that the homotopy groups of {\Omega  X} are the appropriate type. Admittedly, however, we wanted Eilenberg-MacLane spaces to be CW complexes. I am now implicitly invoking a big theorem of Milnor that the loop space of a CW complex has the homotopy type of a CW complex.

So this means that the family of spaces {K(G, n)_{n \geq  0}} for some fixed {G} has maps {\Sigma K(G,n) \rightarrow K(G, n+1)}. This is a general phenomenon, and such families are called spectra; they arise as the representing objects of generalized cohomology theories (in our case, usual cohomology). The fact that the adjoints {K(G,n)  \rightarrow \Omega K(G,n+1)} are isomorphisms turns out to be a manifestation of the suspension isomorphisms {H^i(X)  \simeq H^i(\Sigma X)}.



Finally, we shall actually show how one can get the {K(G,n)}‘s without any fancy representability machinery. In fact, we prove:


Theorem 4 Let {G, n} be such that {G} is abelian. Then {K(G,n)} exists.

Proof: We shall construct this in a very explicit way.

The first thing to do is to note that there is a free presentation

\displaystyle  0 \rightarrow F' \rightarrow F \rightarrow G \rightarrow 0

with {F', F} free on generating sets {I,  J}. We can explicit think of this is as represented a giant matrix {\{a_{ij}\}_{i \in I, j \in J}}. Now we want to turn this gigantic matrix into a map of spaces.

To do this, we shall consider the wedges {X = \bigvee_{I}  S^n} and {Y = \bigvee_J S^n}. The claim is that the homotopy groups in dimension {n} are just {F' = \mathbb{Z}[I]} and {F =  \mathbb{Z}[J]} respectively. Since homotopy groups commute with filtered colimits (because the spheres are compact), we just need to prove it for a finite set.


Lemma 5 Let {I} be a finite set. Then the {I}-indexed inclusions {S^n \rightarrow \bigvee_I S^n} generate {\pi_n(\bigvee S^n)} as a free abelian group.


We want to show that {\pi_n(\bigvee_I S^n) \simeq  \mathbb{Z}[I]}. The easiest way to see this is probably the Hurewicz theorem. We know that {H_n(\bigvee_I  S^n)} is {\mathbb{Z}[I]} with generators represented by the images of a generator of {S^n} under each of the {|I|} inclusions {S^n \rightarrow \bigvee_I S^n}, and the homology groups are zero below {n}. The Hurewicz theorem then implies the result.

Since we know that the homology of Let’s postpone this lemma for the moment. So we know that the map on the homotopy groups is {F  \rightarrow F'} that we want is describable by a matrix {a_{ij} }. So we construct, for {i \in  I}, a map from the {i}th copy of {S^n}, {S^n \rightarrow \bigvee  S^n} that induces {a_{ij}} times a generator for {j \in J}.

When we splice this together, we get a map

\displaystyle  X \rightarrow Y

whose map on {\pi_n} is just {F  \rightarrow F'}. Let us consider the mapping cone of {X \rightarrow Y}, which we call {Z}. The claim is that {Z} has the appropriate homotopy group in dimension {n}.

There is an exact sequence in homology

\displaystyle  \dots \rightarrow H_k(X) \rightarrow H_k(Y) \rightarrow  H_k(Z) \rightarrow H_{k-1}(X) \rightarrow \dots

which can be obtained by converting {X \rightarrow  Y} into a cofibration {X' \rightarrow  Y}, whence the above exact sequence is just the usual exact sequence of a pair in homology {(Y, X')} together with the fact that homology of {(Y,X')} is reduced homology of {Y/X'}.

This long exact sequence and the known homology of {X,Y} (which are wedges of spheres) shows that the homology of {Z} is trivial in dimension {<n}. In dimension {n}, the long exact sequence shows that {H_n(Z) \simeq F}. Now, the Hurewicz theorem states that {\pi_n(Z) \simeq  F}, since the lowest homology and homotopy groups are isomorphic for a simply connected space.

So we have a space, which I’ll call {K_n}, such that {\pi_n(K_n) \simeq F}. Since {K_n} by construction has no cells in dimension {<n}, we already have that {\pi_k(K_n) =  0} for {k<n}. We don’t know anything about the higher homotopy groups of {K_n}, though; the homotopy groups of spheres are far from trivial. We have to do something else.

The general procedure involves adding additional cells to destroy the remaining homotopy groups. Let us state this as a lemma.


Lemma 6 (Homotopy riddance) Let {X} be a pointed space and {m} an integer. There is an inclusion {X \hookrightarrow  X'} such that {X'} is obtained by attaching {m+1}-cells from {X} and {\pi_m(X') = 0}.


This proof is immediately going to imply the existence of Eilenberg-MacLane spaces. Indeed, given {K_n}, we attach {n+2}-cells to eliminate the homotopy groups in dimension {n+1}. This doesn’t change the homotopy groups in dimension {\leq n} by the cellular approximation lemma. So we get a new CW complex {K_{n}'}.

Then we eliminate the homotopy groups of {K_n'} in dimension {n+2} to get a new {K_n''}. When we take the direct limit of these, we get a CW complex which is the Eilenberg-MacLane space.

So let us prove the homotopy riddance lemma. I don’t think it is usually called that. Anyway, let {X} be the space in question; we can look at a set {A} of maps {S^n \rightarrow X} such that every map is homotopic to one of that form. We use these to attach {n+1}-cells to get a new space {X'}. Any map {S^n \rightarrow X'} can be homotoped (by cellular approximation) to have image in {X  \subset X'}, so it is homotopic to something map in {A}, and thus becomes homotopically trivial in {X'}. This completes the proof of the lemma.