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](https://pi.math.cornell.edu/~hatcher/SSAT/SSch1.pdf), 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?



Calculation of $H^{10}(K(\mathbb{Z}, 3); \mathbb{Z})$
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 $2$.


