In Hatcher's Chapter 5 ([https://pi.math.cornell.edu/~hatcher/AT/ATch5.pdf](https://pi.math.cornell.edu/%7Ehatcher/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?



Why does $\iota_4^2 \in H^8(K(\mathbb Z/2,4);\mathbb Z/2)$ not come from $H^8(K(\mathbb Z/2,4);\mathbb Z)$?
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**](http://dx.doi.org/10.1112/blms/bds090), Bull. Lond. Math. Soc. 45, No. 2, 329-340 (2013). [ZBL1270.57068](https://zbmath.org/?q=an:1270.57068)


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**](http://dx.doi.org/10.2307/1970305), Ann. Math. (2) 74, 24-51 (1961). [ZBL0112.14501](https://zbmath.org/?q=an:0112.14501).


