Multiplicative structure on spectral sequence

This is an expansion of John Rognes' answer. I have filled in a few details in Douady's seminare notes and noticed that one gets away with slightly weaker axioms. If there is already a reliable reference for all this, please let me know.

Recall that a Cartan-Eilenberg system $(H,eta,partial)$ (see here, chapter XV.7) consists of modules $H(p,q)$ for each $ple q$, morphisms $etacolon H(p',q')to H(p,q)$ for all $ple p'$, $qle q'$, and boundary morphisms $partialcolon H(p,q) to H(q,r)$ for all $ple qle r$, such that

1. $eta=mathrm{id}colon H(p,q)to H(p,q)$,




2. $eta=etacircetacolon H(p'',q'')to H(p',q')to H(p,q)$,




3. $eta$ and $partial$ commute,




4. there are long exact sequences $cdotsto H(q,r)stackreletato H(p,r)stackreletato H(p,q)stackrelpartialto H(q,r)tocdots$.

The conditions needed for convergence have been omitted. A typical example is $H(p,q)=tilde h_bullet(X_p/X_q)$, where $cdotssupset X_{-1}supset X_0supset X_1supsetcdots$ is a decreasing sequence of cofibrations and $tilde h_bullet$ is some generalised homology theory. The grading is suppressed in the following, but you can easily fill it in.

To set up a spectral sequence from $(H,eta,partial)$, one defines
$$Z^r_p=mathrm{im}bigl(H(p,p+r)stackreletato H(p,p+1)bigr);,$$
$$B^r_p=mathrm{im}bigl(H(p-r+1,p)stackrelpartialto H(p,p+1)bigr);,$$
$$E^r_p=Z^r_p/B^r_p;,$$
$$d^r_pcolon Z^r_p/B^r_ptwoheadrightarrow Z^r_p/Z^{r+1}_pcong
B^{r+1}_{p+r}/B^r_{p+r}hookrightarrow Z^r_{p+r}/B^r_{p+r};.$$
Details are in Switzer's book, chapter 15. In particular
$$ker(d^r_p)=Z^{r+1}_p/B^r_pqquadtext{and}qquad
mathrm{im}(d^r_p)=B^{r+1}_p/B^r_p;.$$
For $a=eta(a_0)in H(p,p+1)$, $a_0in H(p,p+r)$, one has
$$d^r_p([a])=[partial a_0]in E^r_pqquadtext{with}qquadpartial a_0in H(p+r,p+r+1);.$$

Definition (Douady, II.A)
Let $(H,eta,partial)$, $(H',eta',partial')$ und $(H'',eta'',partial'')$ be Cartan-Eilenberg systems. A spectral product $mucolon(H',partial')times(H'',partial'')to(H,partial)$ is a sequence of maps
$$mu_rcolon H'(m,m+r)otimes H''(n,n+r)to H(m+n,m+n+r)$$
such that for all $m$, $n$, $rge 1$, the following two diagrams commute.
$require{AMScd}$
begin{CD}
H'(m,m+r)otimes H''(n,n+r)@>mu_r>>H(m+n,m+n+r)\
@Veta'oplus Veta''V@VVeta V\
H'(m,m+1)otimes H''(n,n+1)@>mu_1>>H(m+n,m+n+1)rlap{;,}
end{CD}
begin{CD}
H'(m,m+r)otimes H''(n,n+r)@>mu_r>>H(m+n,m+n+r)\
@Vpartial'otimeseta''oplus Veta'otimespartial''V@VVpartial V\
{begin{matrix}H'(m+r,m+r+1)otimes H''(n,n+1)\oplus\H'(m,m+1)otimes H''(n+r,n+r+1)end{matrix}}@>mu_1+mu_1>>H_{p+q-1}(m+n+r,m+n+r+1)rlap{;.}
end{CD}

The first diagram is weaker than in Douady's notes.
The second can be read as a Leibniz rule.

Theorem (Douady, Thm II)
A spectral product $mucolon(H',partial')times(H'',partial'')to(H,partial)$ induces products
$$mu^rcolon E^{prime r}_motimes E^{primeprime r}_nto E^r_{m+n};,$$
such that

1. $mu^1=mu_1$




2. $d^r_{m+n}circmu^r=mu^rcirc(d^{prime r}_motimesmathrm{id})pmmu^rcirc(mathrm{id}circ d^{primeprime r}_n)$,




3. $mu^{r+1}$ is induced by $mu^r$.

Proof.
Assume by induction that $mu^r$ is induced by $mu_1$. In particular,
$$Z^{prime r}_motimes Z^{primeprime r}_nstackrel{mu_1}to Z^r_{m+n};,$$
$$B^{prime r}_motimes Z^{primeprime r}_nstackrel{mu_1}to B^r_{m+n};,$$
$$Z^{prime r}_motimes B^{primeprime r}_nstackrel{mu_1}to B^r_{m+n};.$$
This is clear for $r=1$ if we put $mu^1=mu_1$ because $E^1_p=Z^1_p=H(p,p+1)$
and $B^1_p=0$.

Let $[a]in Z^{prime r}_m$, $[b]in Z^{primeprime r}_n$ be represented by $a=eta'(a_0)in H'(m,m+1)$, $b=eta''(b_0)in H''(n,n+1)$ with $a_0in H'(m,m+r)$, $b_0in H''(n,n+r)$.
Using the first diagram and the construction of $d^r_{m+n}$, we conclude that
$$(d^r_{m+n}circmu^r)([a]otimes[b])=d^r_{m+n}[mu_1(aotimes b)]=d^r_{m+n}[eta(mu_r(a_0otimes b_0))]=(partialcircmu_r)(a_0otimes b_0);.$$
From the second diagram, we get
$$(partialcircmu_r)(a_0otimes b_0)=mu_1(partial'a_0otimeseta''b_0)pmmu_1(eta'a_0otimespartial''b_0)=mu^r(d^{prime r}_m[a]otimes[b])pmmu^r([a]otimes d^{primeprime r}_n[b]);.$$
This proves the Leibniz rule (2).

From the Leipniz rule and the facts that $ker(d^r_p)=Z^{r+1}_p/B^r_p$ and $mathrm{im}(d^r_p)=B^{r+1}_p/B^r_p$, we conclude that $mu^r$ induces a product on $E^{r+1}_pcongker(d^r_p)/mathrm{im}(d^r_p)$, which proves (3). Because $mu^r$ is induced by $mu_1$, so is $mu^{r+1}$, and we can continue the induction.