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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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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... – lentic catachresis Jan 27 '15 at 18:08
I have the same problem. I do not understand why the product $ax_2$ generates $E^{2,1}_2$ – Nimza Apr 17 at 13:22

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