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Aug 11, 2022 at 14:33 comment added user30211 @WilberdvanderKallen I realized it's just $\alpha$ and I had the definition wrong...
Aug 11, 2022 at 6:59 comment added Wilberd van der Kallen @Bob Zinckel But it is not like that. `got to be' does not always work. The first derived couple does not have $D$ itself in it.
Aug 10, 2022 at 23:31 history edited user30211 CC BY-SA 4.0
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Aug 10, 2022 at 22:03 history edited YCor
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Aug 10, 2022 at 21:31 comment added user30211 @A.S. Now, if the triangle is $D \stackrel{\alpha}{\rightarrow} D \rightarrow E \rightarrow D[1]$, then isn't it true that the $n$th page is determined by the fact that one of the elements of the triangle has got to be $D \stackrel{\alpha^n}{\rightarrow} D$?
Aug 10, 2022 at 21:21 comment added user164898 If two of the three objects in your exact triangle coincide, then you can build an exact couple. But if all three are pairwise distinct, there's no clear way that an exact couple gets associated to the exact triangle.
Aug 10, 2022 at 21:12 history asked user30211 CC BY-SA 4.0