Ring structure in the Serre spectral sequence

I've tried to understand what's going on in Example 1.5 on page 27-28 in Hatcher's notes on spectral sequences. There is one part in the reasoning that I can't understand here. He writes down a table with the $E^2$-page of a spectral sequence looking like (arrows omitted)

$$\begin{array}{ccccccc} \mathbb{Z}a & 0 & \mathbb{Z}ax_2 & 0 & \mathbb{Z}ax_4 & 0 & \ldots\\ \mathbb{Z}1 & 0 & \mathbb{Z}x_2 & 0 & \mathbb{Z}x_4 & 0 &\ldots \end{array}$$

What I'm trying to figure out is how do we know that $ax_2$ is the generator of $E^{2,1}_2$? Hatcher just writes:

The generators for the $\mathbb{Z}$'s in the upper row are $a$ times the generators in the lower row, because the product $E_2^{0,q}\times E_2^{s,t}\to E_2^{s,t+q}$ is just multiplication of coefficients.

Can someone explain to me what's going on here?

• Remember that the $E_2$ pages has $E_2^{pq} = H^p(B; H^qF)$. Assuming the system of coefficients is simple (which I think it is here), then you can use the universal coefficient theorem to show that (in this case!) $E_2^{pq} = E_2^{p0} \otimes E_2^{0q}$. So this tells you, at least, you should have a copy of $\mathbb{Z}$ in $E_2^{2,1}$. The multiplicative structure is part of the statement of the theorem for the cohomology spectral sequence... To see why that's true you'd have to open up the black box a little bit. – Dylan Wilson Apr 29 '11 at 21:39
• For what it's worth, I had the exact same problem, googled it and ended up here. I've been thinking about this on and off for some days, sometimes I think I've convinced myself that it's true, but then I realize it's not the case... – Bruno Stonek Jan 27 '15 at 18:08
• I have the same problem. I do not understand why the product $ax_2$ generates $E^{2,1}_2$ – Appliqué Apr 17 '16 at 13:22

The OP refers to the Serre spectral sequence in cohomology for the path-loop fibration $$K(\mathbb{Z},1) \to P \to K(\mathbb{Z},2)$$. In general for a fibration $$F \to E \to B$$, the $$E^2$$ page is given by $$E_2^{pq} = H^p(B,H^q(F))$$, where we view $$H^q(F)$$ as a local system under the monodromy action. But in this case, $$K(\mathbb{Z},2)$$ is simply connected, so there is no monodromy. Even better, as $$K(\mathbb{Z},1) \simeq S^1$$, the fiber $$K(\mathbb{Z},1)$$ has free cohomology groups, namely $$H^q(K(\mathbb{Z},1)) = \begin{cases} \mathbb{Z} & q = 0,1 \\ 0 & \text{else.}\end{cases}$$ Thus the universal coefficient theorem gives $$E_2^{pq} = H^p(K(\mathbb{Z},2)) \otimes H^q(K(\mathbb{Z},1)).$$ Hence, once $$E_2^{2,0} = H^p(K(\mathbb{Z},2))\otimes \mathbb{Z}$$ is determined, by arguing that $$E_3 = E_\infty$$, we immediately know $$E_2^{2,1}$$ as well.
The extra magic of this computation is that the differentials in the Serre spectral sequence are derivations with respect to the ring structure on $$E_2^{pq}$$, which itself is the tensor product of the ring structures on the cohomology of base and fiber (when there is no monodromy).
• Perhaps a better way to say it is this: if my coefficients are a direct sum $A \oplus A’$, then $H^n(X; A \oplus A’) = H^n(X;A) \oplus H^n(X;A’)$. This follows from UCT or from any definition of cohomology via a chain complex. The claim about tensors then follows: for example $H^n(X;\mathbb{Z}) \oplus H^n(X;\mathbb{Z}) = H^n(X; \mathbb{Z}) \tensor \mathbb{Z}^2$. – Joshua Mundinger Dec 6 '18 at 16:01