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I'm totally ignorant about the Universal Coefficient Spectral Sequence (I used to work only with principal ideal domains, where the Universal Coefficient Theorem only amounts to a short exact sequence) but I need now to understand it. The book I was redirected to is the infamous User's Guide to SS where I have only been able to find very general versions (one about spectra and the other with functors on general abelian categories). I guess that an exposition in the simple case of chain complexes of A-modules exists somewhere, but I don't know where to look.

Do you have any suggestion?

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What universal coefficient theorem? A very general theorem which you can use to obtain such results in the hypertor spectral sequence, which you can find explained in Wiebel, or McCleary, among others. – Mariano Suárez-Alvarez Feb 4 '11 at 20:51 has a reference to Rotman. Künneth theorems are generalizations of universal coefficient ones. – lentic catachresis Nov 26 '15 at 18:40
Essentially, theorem 10.90 in Rotman's Intro to homological algebra, second edition. – lentic catachresis Nov 26 '15 at 18:59

Well, I'm not sure that my answer is what you want.

I learned the universal coefficient spectral sequence from the following two sources.

J. Levine, "Knot modules I" (1977, Tran. AMS) Cochran-Orr-Teichner "Knot Concordance, Whitney Towers and L2-Signatures" (2003, Ann. Math.)

In Levine's paper (see Thm. 2.3 in page 5), Levine gives a quick and readable construction of UCSS.

In Cochran-Orr-Teichner's one (see Remark. 2.8, proof of Thm. 2.13), they use UCSS to control the dimension of homology groups with coefficient in (noncommutative) quotient field of Ore domain. Also, they gives some examples (Reamrk 2.8) about collapsing conditions (you might want) of UCSS.

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Switzer sends the interested reader to Adams' Lectures on generalised cohomology and Stable homotopy and generalised homology. I couldn't promise that these don't deal with spectra, but at least they'll be geared towards topology... Why are you trying to avoid spectra, by the way?

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