Timeline for Does homology have a coproduct?
Current License: CC BY-SA 3.0
7 events
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| Jul 4, 2014 at 4:07 | history | edited | darij grinberg | CC BY-SA 3.0 |
replaced by latex some old unicode that conflicted with markup; please check the maths
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| Oct 21, 2009 at 12:40 | comment | added | Tyler Lawson | You need that singular chains are free to get the conclusion about the Tor spectral sequence in the first place; the spectral sequence is a general one computing H_(C ⊗_R D) from H_ C and H_* D when C and D are (nonnegative) chain complexes of R-modules with one of them levelwise free. One example is to look at the mod-4 homology of RP^2 x RP^2 from the mod-4 homology of its factors. Having said that, the spectral sequences you get always collapse at E_3 because everything is arising from integral coefficients, but if you leave the higher Tors out it doesn't work. | |
| Oct 21, 2009 at 9:02 | comment | added | Andrew Stacey | Also, although this is the most general form for chains, for singular cohomology then it's a little elaborate, isn't it? After all, singular chains are free (by definition!) so the complication of coefficients doesn't arise. Or am I missing something? | |
| Oct 16, 2009 at 12:02 | history | edited | Tyler Lawson | CC BY-SA 2.5 |
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| Oct 16, 2009 at 12:01 | comment | added | Tyler Lawson | yargh, I meant the Kunneth formula. fixed. | |
| Oct 16, 2009 at 8:21 | comment | added | Andrew Stacey | The universal coefficient theorem wasn't stated in a previous answer. | |
| Oct 15, 2009 at 11:58 | history | answered | Tyler Lawson | CC BY-SA 2.5 |