TITLE: How to compute $[CP^2, G/PL]$?
QUESTION [8 upvotes]: 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.

REPLY [7 votes]: 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 $\mathbb Z$.