How to compute $[CP^2, G/PL]$?

Question

Let $E^4$ be the two stage Postnikov space appearing in the homotopy type of the classifying space $G/PL$. One of its properties is that it only has two nontrivial homotopy groups $\pi_2(E)=Z/2Z$ and $\pi_4(E)=Z$. Here $Z$ denotes the integers.

In Lopez de Medrano's book on involutions, it is stated without proof that the homotopy classes of maps $[CP^2, G/PL]$ is isomorphic to $Z$.

Anyone know a proof? He gives a passing hint that it has to do with the k-invariant used to glue the Postnikov pieces together.

Thanks.

Answer 1

In general, suppose we want to compute homotopy classes $\pi_0 [X, Y]$ of maps $X \to Y$. Assume for simplicity that $Y$ is simply connected, which is the case here. Then we can use the Postnikov tower of $Y$; write the $n$-truncated part of it as $Y_n$. If we know $\pi_0 [X, Y_n]$, we can compute $\pi_0 [X, Y_{n+1}]$ by taking each element of $\pi_0 [X, Y_n]$ and computing how it lifts to $Y_{n+1}$. There is a fiber sequence

$$B^{n+1} \pi_{n+1} Y \to Y_{n+1} \to Y_n$$

which, given that $Y$ is simply connected, is principal; this means that the obstruction to lifting a map $X \to Y_n$ to a map $X \to Y_{n+1}$ is a cohomology class in $H^{n+2}(X, \pi_{n+1} Y)$, the pullback of the Postnikov invariant in $H^{n+2}(Y_n, \pi_{n+1} Y)$ classifying the above fiber sequence. When this obstruction vanishes, the set of lifts is classified by $H^{n+1}(X, \pi_{n+1} Y)$. In particular, if $X$ is a $d$-dimensional CW complex and $n \ge d$, then the obstruction classes always vanish and lifts are always unique, which gives

$$\pi_0 [X, Y] \cong \pi_0 [X, Y_n].$$

In our case, $X = \mathbb{CP}^2$ is a $4$-dimensional CW complex and so we only need to go up to $n = 4$. We have $Y_2 \cong B^2 \mathbb{Z}_2$, so

$$\pi_0 [X, Y_2] \cong H^2(\mathbb{CP}^2, \mathbb{Z}_2) \cong \mathbb{Z}_2$$

by universal coefficients. The same is true of $Y_3 \cong Y_2$. The obstruction to lifting to $Y_4$ is a class in $H^5(\mathbb{CP}^2, \mathbb{Z})$, which again vanishes. The set of lifts is $H^4(\mathbb{CP}^2, \mathbb{Z}) \cong \mathbb{Z}$, and we don't need to go beyond this. This gives

$$\pi_0 [X, Y_4] \cong \pi_0 [X, Y] \cong \mathbb{Z} \times \mathbb{Z}_2$$

as sets; I assume the claim is that we should in fact get $\mathbb{Z}$ the abelian group using the infinite loop space structure on $G/PL$, but these computations aren't enough to get at that. With some refinement I believe the above computations show that, as a group, $\pi_0 [X, Y]$ fits into a short exact sequence

$$0 \to \mathbb{Z} \to \pi_0 [X, Y] \to \mathbb{Z}_2 \to 0$$

but I don't know how to push them to distinguish $\mathbb{Z}$ from $\mathbb{Z} \times \mathbb{Z}_2$.

Answer 2

This is the second half of an answer completing the first half given by @Qiaochu Yuan.

The information we need is that the $k$-invariant is $\beta Sq^2$, and the space $E$ has the $H$-space structure of the fiber of $\beta Sq^2: K(\mathbb Z/2,2)\to K(\mathbb Z,5)$. (Madsen-Milgram Theorem 4.34, but they do not prove this, only refer to Sullivan.)

Since all maps between the EM-spaces are $H$-maps, the short exact sequence one obtains is a short exact sequence of groups.

Let $F$ be the fiber of $\beta: K(\mathbb Z/2,4)\to K(\mathbb Z,5)$, in fact $F=K(\mathbb Z/2,4)$.

The fiber sequence $K(\mathbb Z,4)\to F \to K(\mathbb Z/2,4)$ gives rise to another short exact sequence when applying $[\mathbb C P^2,-]$ to it, namely $0\to \mathbb Z \to \mathbb Z \to \mathbb Z/2\to 0$. Now $Sq^2$ induces a map of the short exact sequences which is the identity on the $\mathbb Z$-terms on the left and also the identity on the $\mathbb Z/2$ terms on the right, since $Sq^2:H^2(\mathbb C P^2,\mathbb Z/2)\to H^4(\mathbb C P^2,\mathbb Z/2)$ is an isomorphism. It follows that the group we are interested in is isomorphic to $\mathbb Z$.