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Aug 18, 2020 at 8:33 comment added Jeremy Rickard Crossposted on MSE
Aug 18, 2020 at 5:12 comment added Aaron Since $Z$ has $0$ differential, the differential of $Z\otimes C$ is $(-1)^a_Z\otimes \Delta_C$, and since everything is flat, you should get $H(Z\otimes C)\cong Z\otimes H(C)$. I cannot figure out what else might be going on.
Aug 18, 2020 at 3:44 history edited user163897 CC BY-SA 4.0
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Aug 18, 2020 at 3:26 comment added user163897 OK. I really appreciate it!
Aug 18, 2020 at 3:25 history edited user163897 CC BY-SA 4.0
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Aug 18, 2020 at 3:20 comment added Yemon Choi Just to let you know it is 4am here so I am unlikely to respond for some time ...
Aug 18, 2020 at 3:16 comment added user163897 yes! it is the proof of his theorem! I will add all from the beginning up to this point! There's not much! Thx again!
Aug 18, 2020 at 3:12 comment added Yemon Choi Something doesn't look right here: if I start with A already having zero differential, so that A=Z, then the claim you mention seems to assert that the homology of the total complex $A\otimes C$ doesn't depend on the differential of $C$. Since I don't have a copy of Rotman's book at hand, could you add some of the surrounding context? Is this something to do with a K\"unneth formula?
Aug 18, 2020 at 3:05 comment added user163897 Yes! It is from page 679 of Rotman's introduction to homological algebra. Thx!
Aug 18, 2020 at 3:03 comment added Yemon Choi Could you tell us which source you are using for this notation/terminology? I am not sure what a "cycle subcomplex" of $A$ is supposed to be: do you mean that you are taking $Z_n$ to be the kernel of $A_n \to A_{n-1}$, and then viewing $Z_{\bullet}$ as a complex with zero differential?
Aug 18, 2020 at 2:50 review First posts
Aug 18, 2020 at 9:41
Aug 18, 2020 at 2:50 history asked user163897 CC BY-SA 4.0