Computing H_2 from pi_1=Z and pi_2

(Related question: What part of the fundamental group is captured by the second homology group?)

Suppose I have a path-connected space $X$ for which $\pi_1(X)=\mathbb{Z}$. Suppose I know $\pi_2(X)$ as a $\pi_1(X)$-module, and I want to compute $H_2(X)=H_2(X;\mathbb{Z})$.

Claim: $H_2(X)$ is isomorphic to the largest quotient of $\pi_2(X)$ on which $\pi_1(X)$ acts trivially (ie the group of coinvariants).

One can prove this by passing to universal covers (assuming $X$ is locally path-connected, blah, blah, blah). The Cartan-Leray spectral sequence of the regular cover $\tilde{X}\to X$ degenerates at the $E_2$ page (since $K(\mathbb{Z},1)$ is a circle) and it follows that $$H_2(X)\cong H_0(\mathbb{Z};H_2(\tilde{X}))\cong H_0(\mathbb{Z};\pi_2(\tilde{X}))\cong H_0(\mathbb{Z};\pi_2(X)).$$ This tells me the answer, but teaches me nothing.

Question: Is there a more elementary proof of this fact? (I am prepared to accept that "elementary" might still involve Postnikov towers, obstruction theory, etc)

Edit: Many thanks for all the helpful answers, I wasn't expecting four alternatives! A remark and a question:

1. This is true with an arbitrary free group replacing $\mathbb{Z}$, and
2. it seems to be well known. Has anyone seen it written down anywhere?
• Hatcher has an exercise leading to a different proof: p.390, ex. 23. Feb 6 '11 at 20:47
• @Martin, The conclusion of this exercise in the case $n=1$ just says that the Hurewicz map $h$ is surjective. Are you saying that the suggested proof leads to a description of the kernel of $h$? Feb 6 '11 at 22:55
• @Mark: One can consider the pair $(K(\mathbb Z,1),X)$ as in the exercise, and then apply Theorem 4.37 and the long exact sequences in homology and homotopy. Feb 7 '11 at 0:17
• Mark, about your question (2): as you may well know, the fact that Martin O referred to, that $\pi_2(X)\to H_2(X)\to H_2(B\pi_1(X))\to 0$ is exact, is a theorem of Hopf 1942, see MR0006510. Brown's book "Cohomology of groups" is a nice reference for that - perhaps he also comments on the kernel of the Hurewicz map? Feb 7 '11 at 16:11
• Yes Tim, you're right, it's in Brown - as Exercise II.5.1 and Exercise VII.7.6. Feb 7 '11 at 16:58

Here's one way. Suppose first that one has a Serre fibration $F\hookrightarrow E \to S^1$ with $F$ simply connected. Then $\pi_2(E)=\pi_2(F)$ by the exact sequence of homotopy groups, and $\pi_2(F)=H_2(F)$ by Hurewicz.

The map $H_2(F)\to H_2(E)$ is surjective, and its kernel is the image of $\mathrm{id}-\phi$, where $\phi$ is the action of the generator of $\pi_1(S^1)$ on $H_\ast(F)$. This is by the Wang exact sequence and the vanishing of $H_1(F)$. So $H_2(E)$ is isomorphic to $H_2(F)_{\pi_1(S^1)}$ (coinvariants) and hence to $\pi_2(E)_{\pi_1(E)}$.

If $X$ is (say) a path-connected CW complex with $\pi_1(X)=\mathbb{Z}$ then its universal cover is classified by a map $X\to B\mathbb{Z}=S^1$, and we can convert this by the usual procedure into a Serre fibration $F\hookrightarrow E \to S^1$. The mapping fibre, $F$, is simply connected, and $E$ is homotopy equivalent to $X$. Now apply the previous argument.

How elementary? Here's as low tech as I can think of: Any element of $H_2(X)$ is represented by a map from a closed oriented surface $F$ (any cycle can be replaced by a surface by gluing edges appropriately). Now homotop the 1-skeleton of $F$ into a circle in $X$ representing $\pi_1(X)=Z$. Since the surface relation is a commutator, the restriction to the 1-skeleton of $F$ is nullhomotopic. Hence a homotopy extension argument shows that the map $F\to X$ is homotopic to one which factors through $F/F^{(1)}\cong S^2$. Thus the map $[S^2,X]\to H_2(X)$ is onto. Now you just have to show that $[S^2,X]$ is the quotient of $\pi_2(X)$ by the $\pi_1(X)$ action. This is easy, depending on how you choose to understand this action. $\pi_2(X)\to [S^2,X]$ is onto because $X$ is path connected (and homotopy extension), and if two classes in $\pi_2(X)$ are freely homotopic the free homotopy traces out a loop in $\pi_1(X)$, hence related by the action.

• can one really glue edges all the way to getting a surface? Feb 7 '11 at 4:13
• @Mariano - Two dimensions is special because, no matter how you glue together a set of triangles along their edges, you always get a surface. Feb 7 '11 at 15:36

There is a cofibrant way of showing this, i.e. forgetting about coverings, Postnikov towers, fibrations, spectral sequences, etc. There are very nice and simple algebraic models for low-dimensional homotopy types. The simplest are crossed modules, which are group homomorphisms $$\partial\colon C_2\longrightarrow C_1$$ such that $C_1$ acts on the right of $C_2$ and the following two equations are satisfied:

$$\partial(x_2^{x_1})=x_1^{-1}x_2x_2, \qquad x_2^{\partial(y_2)}=y_2^{-1}x_2y_2.$$

Crossed modules can be regarded as non-abelian chain complexes $C_*$ concentrated in degrees $1$ and $2$. The subscript indicates the degree of each element. Notice that the first equation says that $\partial$ is $C_1$-equivariant if we let $C_1$ act on itself by conjugation.

The homology of $C_*$ is usually regarded as homotopy groups: $$\pi_1C_*=C_1/\partial(C_2),\qquad \pi_2C_*=\ker\partial.$$ Notice that $\pi_1C_*$ acts on the right of $\pi_2C_*$.

The canonical example of a crossed module is the homomorphism $$\partial \pi_2(X,Y)\longrightarrow \pi_1Y$$ associated to any pair of spaces $(X,Y)$. The fundamental crossed module of a connected CW-complex $X$ with $1$-skeleton $X^1$ is $$\partial\colon\pi_2(X,X^1)\longrightarrow \pi_1X^1.$$

To any crossed module $C_*$ we can associate a two-step chain complex $$\cdots\rightarrow 0\rightarrow C_2^{ab}\otimes_{\mathbb{Z}[C_1]}\mathbb{Z}\stackrel{\bar{\partial}}\longrightarrow C_1^{ab}\rightarrow 0\rightarrow \cdots$$ by abelianizing $C_1$ and $C_2$ and killing the action of $C_1$ on $C_2^{ab}$. If $C_*$ is the fundamental crossed module of $X$ then the homology of this chain complex is $H_1(X)$ and $H_2(X)$ in the corresponding degrees.

Now assume $\pi_1(X)\cong\mathbb{Z}$. Then the natural projection $C_1=\pi_1X^1\twoheadrightarrow \pi_1X\cong\mathbb{Z}$ has a section $i\colon \pi_1X\rightarrow \pi_1X^1$. This section gives rise to a homotopy equivalence of crossed modules: $$\begin{array}{rcccl} &\pi_2X&\stackrel{0}\longrightarrow&\pi_1X&\\\ {\text{inclusion}}&\downarrow&&\downarrow&\scriptstyle i\\\ &\pi_2(X,X^1)&\longrightarrow&\pi_1X^1& \end{array}$$ In particular, the chain complexes associated to these two crossed modules are quasi-isomorphic. The chain complex of the upper crossed module, given by the trivial homomorphism $0\colon \pi_2X\rightarrow \pi_1X$, is simply $$\cdots\rightarrow 0\rightarrow \pi_2X\otimes_{\mathbb{Z}[\pi_1X]}\mathbb{Z}\stackrel{0}\longrightarrow (\pi_1X)^{ab}\rightarrow 0\rightarrow \cdots,$$ hence we recover the well-known isomorphism $(\pi_1X)^{ab}=H_1X$ and what we wanted to obtain $\pi_2X\otimes_{\mathbb{Z}[\pi_1X]}\mathbb{Z}=H_2X$.