I'm having trouble with Example 1.K, p.25, of John McCleary's book A User's Guide to Spectral Sequences.
Specifically, I don't understand how he defines the "obvious map" in the second paragraph : $$\text{"Notice that there is an obvious mapping from $A^*$ to $H^*$ since $A^*$ is free."}$$
It seems to me such a mapping would involve arbitrary choices because the isomorphism we are given is between quotients, and that it would be non trivial wether these choices could be made in such a way as to give a map of algebras.
I resolved my confusion : it's a straightforward application of the universal property of the free graded-commutative graded algebra on a graded set of generators. My confusion stemmed from the fact that I hadn't really understood that the filtration on $H^*$ was used to filter each factor $H^n$, and that any indeterminacy in choosing representatives of elements in $$E^{p,q}_0=\frac{F^pH^{n}}{F^{p+1}H^{n}}$$ (where $n=p+q$) would remain within $H^n$ so be of given degree, and especially given parity. Someow I thought that the indeterminacy could cause one to choose representatives in a sum of different $H^m$...