Lemma 13.4.13. Let $\mathcal{D}$ be a pre-triangulated category. Let $I$ be a set.

1. Let $X_ i$, $i \in I$ be a family of objects of $\mathcal{D}$.

1. If $\prod X_ i$ exists, then $(\prod X_ i)[1] = \prod X_ i[1]$.

2. If $\bigoplus X_ i$ exists, then $(\bigoplus X_ i)[1] = \bigoplus X_ i[1]$.

2. Let $X_ i \to Y_ i \to Z_ i \to X_ i[1]$ be a family of distinguished triangles of $\mathcal{D}$.

1. If $\prod X_ i$, $\prod Y_ i$, $\prod Z_ i$ exist, then $\prod X_ i \to \prod Y_ i \to \prod Z_ i \to \prod X_ i[1]$ is a distinguished triangle.

2. If $\bigoplus X_ i$, $\bigoplus Y_ i$, $\bigoplus Z_ i$ exist, then $\bigoplus X_ i \to \bigoplus Y_ i \to \bigoplus Z_ i \to \bigoplus X_ i[1]$ is a distinguished triangle.

Proof. Part (1) is true because $[1]$ is an autoequivalence of $\mathcal{D}$ and because direct sums and products are defined in terms of the category structure. Let us prove (2)(a). Choose a distinguished triangle $\prod X_ i \to \prod Y_ i \to Z \to \prod X_ i[1]$. For each $j$ we can use TR3 to choose a morphism $p_ j : Z \to Z_ j$ fitting into a morphism of distinguished triangles with the projection maps $\prod X_ i \to X_ j$ and $\prod Y_ i \to Y_ j$. Using the definition of products we obtain a map $\prod p_ i : Z \to \prod Z_ i$ fitting into a morphism of triangles from the distinguished triangle to the triangle made out of the products. Observe that the “product” triangle $\prod X_ i \to \prod Y_ i \to \prod Z_ i \to \prod X_ i[1]$ is special in the terminology of Remark 13.4.4 because products of exact sequences of abelian groups are exact. Hence Remark 13.4.4 shows that the morphism of triangles is an isomorphism and we conclude by TR1. The proof of (2)(b) is dual. $\square$

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