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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?

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    $\begingroup$ 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. $\endgroup$ – Dylan Wilson Apr 29 '11 at 21:39
  • $\begingroup$ 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... $\endgroup$ – Bruno Stonek Jan 27 '15 at 18:08
  • $\begingroup$ I have the same problem. I do not understand why the product $ax_2$ generates $E^{2,1}_2$ $\endgroup$ – Appliqué Apr 17 '16 at 13:22
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I'll attempt to add to Dylan Wilson's excellent comment.

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).

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  • $\begingroup$ Could you please explain how universal coefficient theorem gives that identity? Because U. C. T for cohomology does not involve any tensor.. $\endgroup$ – Paladin Dec 6 '18 at 15:55
  • $\begingroup$ 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 $H^n(X;\mathbb{Z}) \oplus H^n(X;\mathbb{Z}) = H^n(X; \mathbb{Z}) \tensor \mathbb{Z}^2$. $\endgroup$ – Joshua Mundinger Dec 6 '18 at 16:01
  • $\begingroup$ Okay Thank you. $\endgroup$ – Paladin Dec 6 '18 at 16:51

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