The Stacks project

Remark 13.9.11. Let $\mathcal{A}$ be an additive category. Let $0 \to A_ i^\bullet \to B_ i^\bullet \to C_ i^\bullet \to 0$, $i = 1, 2$ be termwise split exact sequences. Suppose that $a : A_1^\bullet \to A_2^\bullet $, $b : B_1^\bullet \to B_2^\bullet $, and $c : C_1^\bullet \to C_2^\bullet $ are morphisms of complexes such that

\[ \xymatrix{ A_1^\bullet \ar[d]_ a \ar[r] & B_1^\bullet \ar[r] \ar[d]_ b & C_1^\bullet \ar[d]_ c \\ A_2^\bullet \ar[r] & B_2^\bullet \ar[r] & C_2^\bullet } \]

commutes in $K(\mathcal{A})$. In general, there does not exist a morphism $b' : B_1^\bullet \to B_2^\bullet $ which is homotopic to $b$ such that the diagram above commutes in the category of complexes. Namely, consider Examples, Equation (107.56.0.1). If we could replace the middle map there by a homotopic one such that the diagram commutes, then we would have additivity of traces which we do not.


Comments (3)

Comment #62 by Tim on

I think there is a mistake in the proof. Basically what is done in the proof is the following: first find homotopic to zero such that the left square commutes when is replaced by as in the first part of lemma 8.4, then find homotopic to zero such that the right square commutes when is replaced by as in the second part of lemma 8.4, then let . But then since the left square commutes with and with we must have and . However, as far as I understand, the last equality does not always hold, because we only required that initial squares commute up to homotopy.

I believe the right solution is to first replace by homotopic so that the left square would commute, then the right square would still commute up to another homotopy and then replace by where is (we postcompose with so that the addition will not break the commutativity of the left square). In one step, just set .

Comment #65 by on

OK, this is an actual honest mistake. The lemma is just wrong. Damn! I turned it into a remark explaining why it was wrong. Hopefully this is now correct. Thanks very much for your comment.

Comment #66 by Tim on

Yes, you are right. The problem with my solution is that the projection on the part () is not a morphism of complexes, so composition with it does not commute with formation of homotopy.


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