# Calculation of $H^{10}(K(\mathbb{Z}, 3); \mathbb{Z})$

I was trying to calculate $$H^q(K(\mathbb{Z}, 3); \mathbb{Z})$$ for some $$q$$ with the Serre spectral sequence associated to the fibration $$K(\mathbb{Z}, 2) \to PK(\mathbb{Z}, 3) \simeq * \to K(\mathbb{Z}, 3)$$.

I obtained that: $$H^q(K(\mathbb{Z}, 3)) = \mathbb{Z}, 0, 0, \mathbb{Z}x, 0, 0, \mathbb{Z}_2x^2, 0, \mathbb{Z}_3y, \mathbb{Z}_2x^3, 0, \mathbb{Z}_3xy, \mathbb{Z}_2x^4\oplus\mathbb{Z}_5w.$$ But in Hatcher's book on spectral sequences, chapter 1, he claims that $$H^q(K(\mathbb{Z}, 3)) = \mathbb{Z}, 0, 0, \mathbb{Z}x, 0, 0, \mathbb{Z}_2x^2, 0, \mathbb{Z}_3y, \mathbb{Z}_2x^3, \mathbb{Z}_2z, \mathbb{Z}_3xy, \mathbb{Z}_2x^4\oplus\mathbb{Z}_5w.$$

The only difference is in $$H^{10}(K(\mathbb{Z}, 3))$$, that for me is $$0$$, while for Hatcher is $$\mathbb{Z}_2z$$. I cannot understand why this happens.

My reasoning is: $$H^{10}(K(\mathbb{Z}, 3))$$ is in position $$(10, 0)$$ and it can be reached by groups in position $$(9 - r, r)$$. These are non trivial only if $$r$$ is even and $$9-r = 0, 3, 6$$. The only possibility is then $$r = 6$$, i.e., $$(3, 6) = H^3(K(\mathbb{Z}, 3); H^6(K(\mathbb{Z}, 2))) = H^3(K(\mathbb{Z}, 3); \mathbb{Z}n^3) = \mathbb{Z}n^3x.$$ But my claim is that $$(10, 0)$$ could not be reached neither by $$(3, 6)$$ because this dies turning $$E_2$$. In fact $$d_2(n) = x$$, so $$d_2(n^3x) = 3n^2x^2 = n^2x^2$$ and so $$d_2: (3, 6) \to (6, 4) = \mathbb{Z}_2n^2x^2$$ is an isomorphism (edit: the error is here, it is not an isomorphism, but it has kernel $$2\mathbb{Z}n^3x$$). Then $$(10, 0)$$ would survive at $$\infty$$, which is not possible.

What's wrong with this?

• Typo: the group in dimension 6 is $\mathbb{Z}/2$, not $\mathbb{Z}$. – John Palmieri Jan 3 at 23:59

I do not like naming a cohomology class $$n$$ because that deserves to be the name of an integer. I will use the name Hatcher does and call the generator of $$H^2(K(\Bbb Z, 2); \Bbb Z)$$ by the name "$$a$$".
The map $$d_3: E_3^{0, 8} \to E_3^{3,6}$$ sends $$d_3(a^4) = 4a^3 x$$ by the Leibniz rule. The map $$d_3: E_3^{3,6} \to E_3^{6,4} \cong \Bbb Z_2 a^3 x^2$$ sends $$a^3x$$ to $$a^3x^2$$; that is, this map is reduction mod 2 in this basis. (You already calculated that this must be true earlier in the computation.)
The homology group of $$\Bbb Z \xrightarrow{4} \Bbb Z \xrightarrow{\pmod 2} \Bbb Z_2$$ in the middle is $$\Bbb Z_2$$. Thus $$E_4^{3,6} = E_7^{3,6} = \Bbb Z_2\langle 2a^3 x\rangle,$$ where the angle brackets indicate that the nonzero class came from the element $$2a^3 x$$ on the $$E_3$$ page.
Thus there is indeed something left at $$E_7$$ in this position, so that the differential $$d_7: E_7^{3,6} \to E_7^{10,0}$$ must be an isomorphism.
It seems what you missed is that $$d_3: \Bbb Z = E_3^{0,8} \to E_3^{3,6} = \Bbb Z$$ is $$4$$, not $$2$$.