Show the spectral sequence of a fibre sequence is a spectral sequence of algebras

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I'm reading Notes on the K-theory of Finite Fields by Steve Mitchell. Here is a lemma from this paper.

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We want to show that the maps in the spectral sequence of the fibre sequence is $\mathbb{Z}/l$ -algebra homomorphisms. How do we show this in general? Why do we need to prove that localization by $l$ is a delooping while the original sequence is also a delooping?

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Even if the spaces can be individually delooped, you also need to deloop the maps in the sequence as described in the given proof. In particular, you would want to show that the map $\psi^q - 1: BU \to BU$ is deloopable.

The (double) loop space structure on $BU$ is given by Bott periodicity: we have the Bott map $\beta: BU \xrightarrow{\sim} \Omega^2_0 BU$. To show that $\psi^q - 1$ is a double loop map, we need to find $f$ such that $$\require{AMScd} \begin{CD} BU @>{\psi^q - 1}>> BU \\ @V{\beta}V{\sim}V @V{\sim}V{\beta}V \\ \Omega^2_0 BU @>>{\Omega^2 f}> \Omega^2_0 BU \end{CD}$$

Going right and then down, we have the map $\beta(\psi^q - 1)$, which is equivalent to $(\frac{1}{q} \psi^q - 1)\beta$. So we would like to take $f$ to be $\frac{1}{q} \psi^q - 1$; however, this approach requires dividing by $q$ and so is only possible after localization. I think this is all explained more clearly somewhere earlier in Mitchell's notes.

Once you've shown this and delooped the fiber sequence, we have a fiber sequence of loop spaces (in fact, infinite loop spaces); as you say, the product in homology is given by the concatenation of loop spaces. Consequently, the resulting spectral sequence is one of (commutative) algebras.