(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?
  • 1
    $\begingroup$ Hatcher has an exercise leading to a different proof: p.390, ex. 23. $\endgroup$
    – Martin O
    Feb 6, 2011 at 20:47
  • $\begingroup$ @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 $h$? $\endgroup$
    – Mark Grant
    Feb 6, 2011 at 22:55
  • 2
    $\begingroup$ @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. $\endgroup$
    – Martin O
    Feb 7, 2011 at 0:17
  • 1
    $\begingroup$ 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? $\endgroup$
    – Tim Perutz
    Feb 7, 2011 at 16:11
  • $\begingroup$ Yes Tim, you're right, it's in Brown - as Exercise II.5.1 and Exercise VII.7.6. $\endgroup$
    – Mark Grant
    Feb 7, 2011 at 16:58

3 Answers 3


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.

  • $\begingroup$ can one really glue edges all the way to getting a surface? $\endgroup$ Feb 7, 2011 at 4:13
  • 1
    $\begingroup$ @Mariano - Two dimensions is special because, no matter how you glue together a set of triangles along their edges, you always get a surface. $\endgroup$ Feb 7, 2011 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 $\pi_2X\otimes_{\mathbb{Z}[\pi_1X]}\mathbb{Z}=H_2X$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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