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Apr 12, 2021 at 12:27 vote accept kevkev1695
Apr 12, 2021 at 5:24 comment added Jochen Wengenroth Similar question (without answer): mathoverflow.net/questions/387352/…
Apr 11, 2021 at 21:32 history became hot network question
Apr 11, 2021 at 16:42 answer added Bertram Arnold timeline score: 8
Apr 11, 2021 at 15:50 history edited kevkev1695 CC BY-SA 4.0
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Apr 11, 2021 at 15:41 comment added kevkev1695 Yes, in this case $F^0 $ equals the tensor product again, because it is right-exact.
Apr 11, 2021 at 15:40 comment added Zerox The widely used dual functor of $\text{Ext}$ is $\text{Tor}$, which differs from your definition by changing $\text{Hom}_R(X,-)$ into $-\otimes_{R}X$ and swaps $X$ and $Y$.
Apr 11, 2021 at 15:28 history edited kevkev1695 CC BY-SA 4.0
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Apr 11, 2021 at 15:21 comment added kevkev1695 It should be well-defined, because two different projective resolutions are homotopy equivalent, so applying $\text{Hom}(X,-)$ should yield again two homotopy equivalent chain complexes since the functor is additive.
Apr 11, 2021 at 15:13 comment added John Palmieri Why is this well-defined?
Apr 11, 2021 at 13:31 history asked kevkev1695 CC BY-SA 4.0