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

Question

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?

Answer 1

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 $2$.