## 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.

Q.E.D.

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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### 2 Responses to Fibrations with fiber an Eilenberg-MacLane space

1. jessemckeown says: