Why does $\iota_4^2 \in H^8(K(\mathbb Z/2,4);\mathbb Z/2)$ not come from $H^8(K(\mathbb Z/2,4);\mathbb Z)$?

Question

In Hatcher's Chapter 5 (https://pi.math.cornell.edu/~hatcher/AT/ATch5.pdf) on page 574 (page 57 in the pdf), he states that $\iota_4^2 \in H^8(K(\mathbb Z/2,4);\mathbb Z/2)$ is not in the image of $H^8(K(\mathbb Z/2,4);\mathbb Z)\to H^8(K(\mathbb Z/2,4);\mathbb Z/2)$. The argument for the lower classes ($\iota_4, Sq^2\iota_4, Sq^2Sq^1\iota_4$) which don't come from $H^*(K(\mathbb Z/2,4);\mathbb Z)$ is that the Bockstein homomorphism $Sq^1\colon H^*(K(\mathbb Z/2,4);\mathbb Z/2) \to H^{*+1}(K(\mathbb Z/2,4);\mathbb Z/2)$ applied to those classes is nonzero, so it can't come from $H^*(K(\mathbb Z/2,4);\mathbb Z)$. This argument doesn't work for $\iota_4^2 = Sq^4 \iota_4$ because $Sq^1Sq^4 \iota_4 = Sq^5 \iota_4 = 0$.

How can one argue that $\iota_4^2 \in H^8(K(\mathbb Z/2,4);\mathbb Z/2)$ doesn't come from $H^8(K(\mathbb Z/2,4);\mathbb Z)$? Is it even true?

Answer 1

It is true, and follows from results of Browder on the mod 2 Bockstein spectral sequence for $K(\mathbb{Z}/2,4)$. (We can replace $4$ by any even integer $k$ and conclude that $\iota_k^2$ ia not the reduction of an integral class.) An argument is spelled out in Section 3 of

Grant, Mark; Szűcs, András, On realizing homology classes by maps of restricted complexity, Bull. Lond. Math. Soc. 45, No. 2, 329-340 (2013). ZBL1270.57068

available in preprint form at https://arxiv.org/abs/1111.0249. It is based on Theorem 5.5 of

Browder, W., Torsion in (H)-spaces, Ann. Math. (2) 74, 24-51 (1961). ZBL0112.14501.