Fibrations with fiber an Eilenberg-MacLane space

One of the fundamental constructions of classical homotopy theory is the Postnikov tower of a space X. In homotopy type theory, this is just its tower of truncations:

X \to \cdots \to \Vert X\Vert_n \to \dots \to \Vert X\Vert_2 \to \Vert X \Vert_1 \to \Vert X \Vert_0

One thing that’s special about this tower is that each map \Vert X \Vert_n \to \Vert X \Vert_{n-1} has a (homotopy) fiber that is an Eilenberg-MacLane space K(\pi_n(X),n). This is easy to see from the long exact sequence. Moreover, when X is a special sort of space, such fibrations are classified by cohomology classes, according to the following theorem:

Theorem: Suppose f:Y\to X has (homotopy) fibers that are all merely isomorphic to K(A,n), for some n\ge 1 and some abelian group A. Then the following are equivalent:

  1. f is the homotopy fiber of a map X\to K(A,n+1).
  2. The induced action of \pi_1(X) (at all basepoints) on A is trivial.

A space for which each map \Vert X \Vert_n \to \Vert X \Vert_{n-1} has this property is called simple, and the resulting maps \Vert X \Vert_{n-1} \to K(\pi_n(X),n+1) are called its k-invariants. Note that they live in the cohomology group H^{n+1}(\Vert X \Vert_{n-1},\pi_n(X)).

The above theorem can be found, for instance, as Lemma 3.4.2 in More Concise Algebraic Topology, where it is proven using the Serre spectral sequence. In this post I want to explain how in homotopy type theory, it has a clean conceptual proof with no need for spectral sequences.

Dan recently posted about the construction of Eilenberg-MacLane spaces. Here we’ll only need the fact that they exist, together with the following:

Lemma: The type \mathrm{Map}_*(K(G,n),K(H,n)) of pointed maps between Eilenberg-MacLane spaces is equivalent to the set \mathrm{Hom}(G,H) of group homomorphisms.

Proof: by induction on n\ge 1. When n=1, the construction of K(G,1) as a HIT gives this fairly immediately. For the induction step, by definition we have K(G,n+1) = \Vert \Sigma K(G,n) \Vert_{n+1}. Since K(H,n+1) is an (n+1)-type, pointed maps K(G,n+1) \to K(H,n+1) are equivalent to pointed maps \Sigma K(G,n) \to K(H,n+1), hence to pointed maps K(G,n) \to \Omega K(H,n+1) = K(H,n). \Box

Now suppose given f:Y\to X as in the above theorem. We may as well assume that Y is \sum_{(x:X)} P(x) for some P:X\to \mathcal{U}, and by assumption P factors through the type \sum_{K:\mathcal{U}} \Vert K = K(A,n)\Vert. Thus, we need to analyze the homotopy type of this space, which I will denote \mathrm{EM}(A,n), the type of Eilenberg-MacLane spaces K(A,n).

This type should not be confused with the type \mathrm{EM}_*(A,n), the type of pointed Eilenberg-MacLane spaces, which is defined as \sum_{K:\mathcal{U}_*} \Vert K = K(A,n)\Vert, with \mathcal{U}_* = \sum_{Z:\mathcal{U}}Z the type of pointed types. Since two connected pointed types are merely equal in \mathcal{U} if and only if they are merely equal in \mathcal{U}_*, we can write \mathrm{EM}_*(A,n) = \sum_{K:\mathrm{EM}(A,n)} K. Therefore, the fiber of the forgetful map \mathrm{EM}_*(A,n) \to \mathrm{EM}(A,n) over the canonical space K(A,n) is K(A,n) itself. On the other hand, by the lemma, \mathrm{EM}_*(A,n) can be identified with K(\mathrm{Aut}(A),1). In other words, we have a fiber sequence

K(A,n) \to K(\mathrm{Aut}(A),1) \to \mathrm{EM}(A,n)

If n>1, the resulting long exact sequence tells us that \pi_1(\mathrm{EM}(A,n)) = \mathrm{Aut}(A) and \pi_{n+1}(\mathrm{EM}(A,n)) = A and all other homotopy groups are 0.

If n=1 we have to do a little more work: we compute that the map K(A,1) \to K(\mathrm{Aut}(A),1) corresponds to the map A \to \mathrm{Aut}(A) which is the action of A on itself by conjugation, so that \pi_1(\mathrm{EM}(A,1)) = \mathrm{Out}(A), the outer automorphism group of A (the quotient of this action), while \pi_2(\mathrm{EM}(A,1)) = Z(A), the center of A (the kernel of this action). Since A is abelian, we have \mathrm{Out}(A) = \mathrm{Aut}(A) and Z(A) = A, so this is the same as the case n>1.

It follows that \Vert \mathrm{EM}(A,n) \Vert_1 = K(\mathrm{Aut}(A),1), and the long exact sequence of the truncation map \mathrm{EM}(A,n) \to \Vert \mathrm{EM}(A,n) \Vert_1 tells us that its fiber is K(A,n+1). That is, we have a fiber sequence

K(A,n+1) \to \mathrm{EM}(A,n) \to K(\mathrm{Aut}(A),1).

It follows that our classifying map P:X\to \mathrm{EM}(A,n) (remember it?) factors through K(A,n+1) if and only if its image in K(\mathrm{Aut}(A),1) is trivial. But K(\mathrm{Aut}(A),1) is a 1-type, so maps from X into it factor uniquely through \Vert X \Vert_1. And a map \Vert X \Vert_1 \to K(\mathrm{Aut}(A),1) essentially consists of maps \pi_1(X) \to \mathrm{Aut}(A) for all basepoints, i.e. actions of \pi_1(X) on A, and is trivial just when those actions are trivial.

We’re almost done with the theorem! The only thing left to show is that when the classifying map P:X\to \mathrm{EM}(A,n) factors through K(A,n+1), then f is in fact the homotopy fiber of the factorization. In other words, we want to show that when the canonical fibration \mathcal{U}_* \to \mathcal{U} is pulled back to K(A,n+1), its domain is contractible. But this pulled back fibration over K(A,n+1) has fiber K(A,n), so contractibility of its domain follows from its long exact sequence.


Note that this proof gives us a bit more: if n=1 and A is not abelian, then f:Y\to X with fiber K(A,1) is classified by a map X\to K(Z(A),2) if and only if \pi_1(X) acts on A through inner automorphisms. But even leaving that slight generalization aside, I find this proof to be more illuminating than a spectral sequence argument. It also makes it clear that if the action is nontrivial, then the k-invariants live in some kind of more general cohomology (specifically, cohomology with local coefficients).

I should emphasize, though, that there is a related surprise for classical homotopy theorists: the space X may not be the limit of its Postnikov tower unless it is already finitely truncated. This is a bit unfortunate, but we can still extract useful information about X from its tower.

For instance, suppose f:X\to Y is a map of simple spaces which induces an isomorphism H^n(Y;A) \cong H^n(X,A) on cohomology with coefficients in all abelian groups A, for all n\ge 0. Now suppose Z is a simple n-type, which is therefore the limit of its Postnikov tower that has k-invariants in ordinary cohomology. By inducting up this tower, therefore, we can show that the induced map [Y,Z] \to [X,Z] is a bijection, where [X,Z] = \pi_0(X\to Z) is the set of homotopy classes of maps.

Now take Z = \Vert X\Vert_n; then the truncation map X\to \Vert X \Vert_n extends to some homotopy class of maps Y\to \Vert X \Vert_n. And by uniqueness, the resulting composite Y\to \Vert X \Vert_n \to \Vert Y \Vert_n must be the truncation map of Y. Applying n-truncation again to this whole sequence of maps X\to Y\to \Vert X \Vert_n \to \Vert Y \Vert_n, we see that \Vert f \Vert_n : \Vert X\Vert_n \to \Vert Y \Vert_n is an equivalence.

In other words, a map of simple spaces which induces an isomorphism on ordinary cohomology is necessarily \infty-connected. This can be extended to “nilpotent” spaces as well, which are basically those whose truncations can be contructed out of Postnikov-like towers with more than one space at each “level”. Thus, for a large class of spaces, isomorphisms on ordinary cohomology suffice to detect isomorphisms of homotopy groups. This gives a bit of a clue as to why algebraic topologists spend so much time computing cohomology.

To get some idea for how amazing this is, consider S^2. This is 1-connected, so it’s simple. It has all sorts of complicated higher homotopy groups, but its cohomology is quite trivial: H^n(S^2;A) = A if n=0,2 and 0 otherwise. This result says that if we have any other simple space Y with the same boring cohomology, H^n(Y;A) = A if n=0,2 and 0 otherwise — and a map S^2 \to Y inducing the isomorphism on cohomology — then Y necessarily has all the same complicated higher homotopy groups as S^2!

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3 Responses to Fibrations with fiber an Eilenberg-MacLane space

  1. jessemckeown says:

    This was fun to read!
    I keep wondering whether “nilpotent” has a generalization that ought to be called “solvable”.

  2. Zara says:

    Greeat read

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