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 (108.62.0.1). 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.

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 $p$ homotopic to zero such that the left square commutes when $b$ is replaced by $b+p$ as in the first part of lemma 8.4, then find $q$ homotopic to zero such that the right square commutes when $b$ is replaced by $b+q$ as in the second part of lemma 8.4, then let $b' = b + p + q$. But then since the left square commutes with $b+p$ and with $b+p+q$ we must have $q \circ \alpha_1 = 0$ and $0 = c \circ \beta_1 \circ \alpha_1 = \beta_2 \circ (b+q) \circ \alpha_1 = \beta_2 \circ b \circ \alpha_1$. 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 $b$ by homotopic $b'$ so that the left square would commute, then the right square would still commute up to another homotopy $g'$ and then replace $b'$ by $b'' = b' + d \circ k + k \circ d$ where $k$ is $s_2 \circ g' \circ s_1 \circ \beta_1$ (we postcompose with $s_1 \circ \beta_1$ so that the addition will not break the commutativity of the left square). In one step, just set $h^n = -f^n \circ \pi_1^n + s_2^{n-1} \circ g^n \circ s_1^n \circ \beta_1^n$.

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 $C_1^\bullet$ part ($s_1 \circ \beta_1$) is not a morphism of complexes, so composition with it does not commute with formation of homotopy.

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