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let us suppose that we are computing homology of a product where none of the requirements of künneth theorem are valid : is there a general way to compute the homology of such products?

Many thanks

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    Which Kunneth theorem do you mean? There are increasingly general versions..2011-05-19
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    What Kunneth formula are you using? There's a general Kunneth formula which holds for products of CW complexes, with coefficients in a PID, which (as I see it) is a fairly inclusive condition. See Hatcher section 3.B for more (the explicit formula is on p275).2011-05-19
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    You should probably describe what kind of situation you have in mind: I am pretty sure there are cases where nothing will work...2011-05-19
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    What coefficients did you have in mind? As long as you're using a commutative ring, there's a Kunneth spectral sequence. If you've never used a SS before, they can be intimidating. If you give more detail, maybe someone here can help further.2011-05-19
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    @Josh: As tempting as it is, please don't use that abbreviation for spectral sequences (at least not in public). It wakes memories to a [very bad era](http://en.wikipedia.org/wiki/Schutzstaffel).2011-05-19
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    @Theo: That never even occurred to me! I apologize and won't use that abbreviation any more.2011-05-19
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    @Josh: No worries, I was not offended at all, no need to apologize. My native language is German, and I'm a bit sensitive to that because we were sort of trained against using abbreviations with such connotations from elementary school on... The mathematical point you're making is a very good one.2011-05-19
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    Thanks to all of you for your comments. I was talking about the basic Künneth formula when the ring is a field or one of the homology module of one the spaces is a free R-module. I was thinking about computing the (co)homology of $RP^{n}xRP^{m}$2011-05-19
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    Well, with $\mathbb{Z}_2$ coefficients, which are the easiest to work with when considering real projective space, you *can* use the "basic" Künneth formula. If you want the integral (co)homology, then see the section in Hatcher I mentioned above. But it'll be more work, and you'll need to learn what Tor is if you don't know already (handily, Hatcher explains all in the preceding section).2011-05-20

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