Triangulated Categories — Examples

This is the canonical biproduct triangle, hence is distinguished.

Its final map is zero, so it also exhibits the splitting criterion and the evident section/retraction.

By [F1], the cone has $D_n\oplus C_{n-1}$ in degree $n$, so only degrees $1$ and $0$ remain and its differential is multiplication by $m$.

By [F2], passing this explicit cone triangle to $K(\mathbb{Z}\text{-Mod})$ makes it a standard cone triangle.

The displayed cone triangle is distinguished.

The representable long exact Hom theorem supplies the exact segment, and $K(S^0\mathbb{Z},S^0\mathbb{Z})\cong \mathbb{Z}$ identifies its first map with multiplication by $m$.

The maps are composable chain maps of stalk complexes.

By [F1] the three-cone calculation provides the comparison maps $\alpha$ and $\beta$. Applying [F2] in $K(\mathbf{A}\mathbf{b})$ supplies the signed fourth face and proves that the resulting triangle is distinguished.

Thickness is the preceding proposition.

A quasi-isomorphism has an acyclic cone, so its standard cone object lies in this thick subcategory.

All consecutive composites are zero because every displayed map is zero.

If it were distinguished, it would be isomorphic to the standard cone triangle of $0:S^0\mathbb{Z}\to 0$, whose final map is the identity of $S^0\mathbb{Z}[1]$; no triangle isomorphism can turn that nonzero map into zero.

The cone is $S^0\mathbb{Z}[1]\oplus S^0\mathbb{Z}[1]$. Besides the zero third component, take the chain map whose matrix has the identity from the second summand to the first and zero in all other entries.

Both third components kill the cone inclusion and are killed by the cone projection, so each completes the same zero square; they differ on the second summand and hence are distinct in $K(\mathbb{Z}\text{-Mod})$.