Lemma 13.4.17. An exact functor of pre-triangulated categories is additive.

Proof. Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor of pre-triangulated categories. Since $(0, 0, 0, 1_0, 1_0, 0)$ is a distinguished triangle of $\mathcal{D}$ the triangle

$(F(0), F(0), F(0), 1_{F(0)}, 1_{F(0)}, F(0))$

is distinguished in $\mathcal{D}'$. This implies that $1_{F(0)} \circ 1_{F(0)}$ is zero, see Lemma 13.4.1. Hence $F(0)$ is the zero object of $\mathcal{D}'$. This also implies that $F$ applied to any zero morphism is zero (since a morphism in an additive category is zero if and only if it factors through the zero object). Next, using that $(X, X \oplus Y, Y, (1, 0), (0, 1), 0)$ is a distinguished triangle by Lemma 13.4.11 part (3), we see that $(F(X), F(X \oplus Y), F(Y), F(1, 0), F(0, 1), 0)$ is one too. This implies that the map $F(X) \oplus F(Y) \to F(X \oplus Y)$ is an isomorphism by Lemma 13.4.11 part (2). To finish we apply Homology, Lemma 12.7.1. $\square$

Comment #7903 by Elías Guisado on

I think the notation may be a little bit confusing: The morphism denoted as $(0,1)$ in $(X, X \oplus Y, Y, (1, 0), (0, 1), 0)$ is not the same morphism as the corresponding one in $F(1, 0) \oplus F(0, 1) : F(X) \oplus F(Y) \to F(X \oplus Y)$. To make the notation unambiguous, I would write $(X, X \oplus Y, Y,\binom{1}{0}, (0, 1), 0)$ for the former expression and $F\binom{1}{0}\oplus F\binom{0}{1}:F(X)\oplus F(Y)\to F(X\oplus Y)$ for the latter one. (This is the notation from https://ncatlab.org/nlab/show/matrix+calculus .)

And about the "we omit the rest of the argument": wouldn't it suffice to invoke 12.7.1?

Comment #7904 by Elías Guisado on

Also: instead of the expression at the beginning shouldn't we write the following expression? I mean, both expressions coincide after we show that $F(0)=0$, but the expression is written before $F(0)=0$ is proven.

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• 13 comment(s) on Section 13.4: Elementary results on triangulated categories

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