Triangulated Categories

TR1 supplies $X\stackrel{1}{\to }X\to 0\to X[1]$; applying TR2 gives the displayed zero triangle.

Distinguished triangles are closed under finite biproducts. Indeed, TR1 completes the biproduct of their first arrows to a distinguished triangle, and TR3 on the two inclusions gives a morphism from the biproduct triangle to this completion that is the identity on its first two objects. For every $W$, the usual TR3 factorization argument on rotations gives exactness of the representable Hom sequence of a distinguished triangle; the sequence of the biproduct triangle is the direct sum of the two exact sequences. The elementary five-lemma diagram chase therefore makes the induced map on third objects bijective after applying $\mathcal{T}(W,-)$ for every $W$, hence an isomorphism by Yoneda. Thus the biproduct triangle is isomorphic to the TR1 completion and is distinguished.

The displayed split triangle is the biproduct of the distinguished identity triangle $X\stackrel{1}{\to }X\to 0\to X[1]$ and the distinguished zero triangle $0\to Y\stackrel{1}{\to }Y\to 0$. Step 2.1 therefore makes it distinguished.

TR2 gives the signed left rotation, and conversely says that its being distinguished is equivalent to that of the original triangle.

Three successive left rotations give the translate of the original triangle by $[1]$, up to the sign automorphisms prescribed by Rotation of a triangle. Hence $T$ is distinguished if and only if $T[1]$ is distinguished.

Iterating this equivalence in both directions gives $T[n]$ for every $n\in \mathbb{Z}$. A right rotation is the inverse of a left rotation up to the same sign isomorphisms, so it too preserves distinguished triangles.

TR1 and TR3 first give $gf=0$. If $u:W\to Y$ satisfies $gu=0$, compare the appropriately rotated identity triangle of $W$ with the given triangle; TR3 supplies $v:W\to X$ with $fv=u$. Rotating the given triangle gives the same factorisation at every translated term.

Thus $\mathcal{T}(W,-)$ is homological. The dual rotated-identity-triangle argument gives exactness for $\mathcal{T}(-,W)$ with the arrows reversed, so it is cohomological.

The two representable functors have the required three-term exactness on the given triangle.

Apply that exactness to every translate, using signed rotations to identify consecutive three-term portions; concatenating them gives exactly the two displayed long sequences.

Apply $\mathcal{T}(W,-)$ to the morphism and use the long exact Hom sequences; the ordinary five lemma makes the map on $\mathcal{T}(W,-)$ induced by the third component an isomorphism for every $W$.

Taking $W$ to be each source and evaluating inverse natural maps at identities yields a two-sided inverse for the third component, hence it is an isomorphism.

Rotate source and target triangles simultaneously until the two known components are adjacent to the unknown one.

The rotated morphism has the same component isomorphisms up to translation, so the triangulated five lemma makes the remaining component invertible; undoing the rotation proves the claim.

If $a=0$, composition with $a$ is zero at every object, so the natural transformation vanishes.

Conversely its component at $X$ sends $1_X$ to $a$; if that component is zero then $a=0$.

In the left rotation $Y\stackrel{g}{\to }Z\stackrel{h}{\to }X[1]\to Y[1]$, the final arrow is zero. Exactness of $\mathcal{T}(X[1],-)$ therefore gives $s:X[1]\to Z$ with $hs=1_{X[1]}$.

Exactness of $\mathcal{T}(Z,-)$ factors $1_Z-sh$ as $gt$ for some $t:Z\to Y$; exactness of $\mathcal{T}(Y,-)$ and $f=0$ give $tg=1_Y$. Put $t^{\mathrm{\prime }}=t-tsh\mathrm{.}$ Since consecutive triangle maps compose to zero, $hg=0$; since $hs=1$, one obtains $t^{\mathrm{\prime }}g=1$, $g{t}^{\mathrm{\prime }}=1-sh$, and $t^{\mathrm{\prime }}s=0$.

The identities in step 2.1 show that $(t^{\mathrm{\prime }},h):Z\to Y\oplus X[1]$ and $(g,s):Y\oplus X[1]\to Z$ are inverse. They intertwine all three displayed maps, so they give the claimed isomorphism of triangles.

If one map vanishes, rotate until it is first and apply A distinguished triangle with zero first map is split; undoing that rotation makes the original triangle split up to rotation.

A canonical biproduct triangle has zero final map. Each of its rotations therefore has one zero map, and this property is preserved by an isomorphism of triangles; this proves the converse.

TR3 extends the identity square on $f$ to a morphism between the two triangles.

Its first two components are identities, so the two-isomorphisms proposition makes its third component an isomorphism; TR3 supplies no uniqueness for that component.

Apply TR4 to precisely these three completions.

Its fourth face is the displayed distinguished triangle; changing a chosen completion only replaces its cone object by a noncanonical isomorphic one.

Naturality at the first two maps gives the first two commuting squares of the image triangles.

The translation-compatibility equation, followed by naturality at the final map, gives the square into $F(X)[1]$; componentwise invertibility makes the resulting morphism an isomorphism of triangles.

The long exact sequence of every translated distinguished triangle makes the total kernel closed under shifts and two-out-of-three.

If $B$ is a retract of $A$ then $H(B[n])$ is a retract of $H(A[n])=0$ for every $n$, hence is zero; this proves direct-summand closure.

A cone long exact sequence and the shift identification show that acyclicity is stable under shifts and satisfies two-out-of-three for cone triangles.

Homology takes a retract of a complex to a retract of its homology, and degreewise finite biproducts preserve acyclicity; hence the class is closed under direct summands and is thick.

The standard cone triangle of $f$ completes $f$, and the definition closes this class under isomorphism.

The cone of $1_C$ is contractible and hence isomorphic to zero in $K(\mathcal{A})$, so its standard triangle gives the TR1 identity triangle.

Write the standard cone triangle as $C\stackrel{f}{\to }D\stackrel{j}{\to }\mathrm{Cone}(f)\stackrel{q}{\to }C[1]\mathrm{.}$ A direct cone calculation for $j$ splits $\mathrm{Cone}(j)$ as $C[1]$ together with the contractible cone of $1_D$. Projection onto $C[1]$ is therefore a homotopy equivalence.

Under that projection the standard triangle of $j$ becomes $D\stackrel{j}{\to }\mathrm{Cone}(f)\stackrel{q}{\to }C[1]\stackrel{-f[1]}{\to }D[1];$ the minus sign is the one forced by the cone differential. Thus the left rotation of every standard cone triangle is distinguished.

The converse uses the right rotation, not merely the inverse rotation of the particular triangle in step 2.1. A second direct cone calculation splits $\mathrm{Cone}(-q[-1])\simeq D\oplus \mathrm{Cone}(1_C)$ compatibly with the canonical maps. Since the identity cone is contractible, the standard cone triangle of $-q[-1]$ is therefore isomorphic in $K(\mathcal{A})$ to $\mathrm{Cone}(f)[-1]\stackrel{-q[-1]}{\to }C\stackrel{f}{\to }D\stackrel{j}{\to }\mathrm{Cone}(f),$ the declared right rotation of the original standard triangle.

Every distinguished cone triangle is isomorphic to a standard one, and left and right rotation preserve isomorphisms of triangles. Steps 2.1 and 3.1 therefore prove both directions of the TR2 equivalence for an arbitrary triangle.

Choose chain-map representatives $a:C\to C^{\mathrm{\prime }}$ and $b:D\to D^{\mathrm{\prime }}$ for the two vertical arrows. Commutativity in $K(\mathcal{A})$ means that there is a chain homotopy $H$ with $bf-f^{\mathrm{\prime }}a=dH+Hd\mathrm{.}$ With the cone convention of Distinguished cone triangle in the homotopy category, the matrix $\left(\begin{array}{cc}b& H\\ 0& a[1]\end{array}\right):\mathrm{Cone}(f)⟶\mathrm{Cone}(f^{\mathrm{\prime }})$ is a chain map.

Its composites with the cone inclusion and projection agree in $K(\mathcal{A})$ with the prescribed vertical arrows (the possible $H$-terms are precisely homotopies). Thus its homotopy class is the required third component of a morphism of triangles; isomorphic replacements of standard cone triangles preserve the conclusion.

Let $\alpha :\mathrm{Cone}(f)⟶\mathrm{Cone}(gf),\beta :\mathrm{Cone}(gf)⟶\mathrm{Cone}(g)$ be the cone maps induced by the squares $(1_C,g)$ and $(f,1_E)$. In the coordinates of The three-cone calculation for a composite chain map, write an element of $\mathrm{Cone}(\alpha )$ as $(e,c,d,c^{\mathrm{\prime }})$ and define $r(e,c,d,c^{\mathrm{\prime }})=(e,d+f(c)),s(e,d)=(e,0,d,0)\mathrm{.}$ The displayed cone-differential calculation in that lemma shows that $r$ and $s$ are chain maps, $rs=1$, and $sr\simeq 1$: under its isomorphism $\mathrm{\Theta }$, $r$ is projection off the contractible $\mathrm{Cone}(1_{C[1]})$ summand and $s$ is inclusion of the $\mathrm{Cone}(g)$ summand.

If $j_{\alpha }$ and $q_{\alpha }$ are the canonical maps in the standard cone triangle of $\alpha$, then the formulas give $r{j}_{\alpha }=\beta ,q_{\alpha }s=j_f[1]\circ q_g\mathrm{.}$ Indeed, $j_{\alpha }(e,c)=(e,c,0,0)$ and $\beta (e,c)=(e,f(c))$, while $q_{\alpha }(e,c,d,c^{\mathrm{\prime }})=(d,c^{\mathrm{\prime }})$ and $j_f[1]q_g(e,d)=(d,0)$.

Thus the standard distinguished cone triangle of $\alpha$ is isomorphic in $K(\mathcal{A})$ to $\mathrm{Cone}(f)\stackrel{\alpha }{\to }\mathrm{Cone}(gf)\stackrel{\beta }{\to }\mathrm{Cone}(g)\stackrel{j_f[1]\circ q_g}{\to }\mathrm{Cone}(f)[1]\mathrm{.}$ The definitions of $\alpha$ and $\beta$ make the other two faces the required morphisms of cone triangles, and the displayed final map is precisely the typed fourth arrow in TR4.

$K(\mathcal{A})$ is additive and its shift is an additive autoequivalence.

The chosen class of cone triangles satisfies TR1, TR2 with the declared sign, TR3, and TR4 by the four preceding lemmas; these data therefore meet the definition of a triangulated category.

The cone long exact sequence identifies $H_n(C)\to H_n(D)\to H_n(\mathrm{Cone}(f))$ as an exact sequence, and homology factors through $K(\mathcal{A})$.

By definition, a distinguished triangle is isomorphic to a standard cone triangle. Functoriality of $H_n$ transports the exact three-term sequence in step 1.1 across such an isomorphism, which is exactly the homological-functor condition.

Applying $F$ degreewise sends chain maps and homotopies to chain maps and homotopies, and preserves the finite biproduct and zero-map formula defining a cone.

Hence $F(\mathrm{Cone}(f))\cong \mathrm{Cone}(Ff)$ compatibly with shift, and it sends each standard cone triangle to one; this is the required exact-functor datum.

The standard cone triangle has third object $\mathrm{Cone}(f)$.

A chain map is a quasi-isomorphism exactly when its cone is acyclic, so this third object is acyclic.

TR3 states only that there exists a third component completing the square.

In $K(\mathbb{Z}\text{-Mod})$, take the standard cone triangle of the zero map $S^0\mathbb{Z}\to S^0\mathbb{Z}[1]$. Its cone is $S^0\mathbb{Z}[1]\oplus S^0\mathbb{Z}[1]$, with the second triangle map the first-summand inclusion and the third triangle map the second-summand projection.

The zero square on the first two terms has the zero third component as one completion. It also has the endomorphism of the cone whose only nonzero matrix entry is the identity from the second summand to the first: this endomorphism kills the inclusion and is killed by the projection, so it completes the same square.

The second endomorphism is nonzero in the homotopy category because it is the identity between stalk summands with zero differentials, while the first completion is zero. Hence the TR3 completion is not unique.

TR1 supplies a completion for each morphism, but its third object is only determined up to nonunique isomorphism.

TR3 likewise supplies, rather than canonically chooses, maps between completions; therefore the axioms do not provide compatible object and morphism choices. An enhancement is extra structure.

Distinguished cone triangles do have zero consecutive composites, but this is only a necessary condition.

In $K(\mathbb{Z}\text{-Mod})$, consider $S^0\mathbb{Z}⟶0⟶S^0\mathbb{Z}[1]⟶S^0\mathbb{Z}[1]$ with all three maps zero. Its consecutive composites vanish.

If this triangle were distinguished, it would be isomorphic to the standard cone triangle of $0:S^0\mathbb{Z}\to 0$. The final map of that cone triangle is the identity of $S^0\mathbb{Z}[1]$; commutativity of the final square would then force the third component of any purported triangle isomorphism to be zero, so it could not be an isomorphism. Thus the displayed triangle is not distinguished.

Associativity already holds in every category, before any distinguished triangles are specified.

TR4 instead compares chosen completions of $f$, $g$, and $gf$ by a fourth distinguished cone triangle, so it supplies genuinely additional triangulated structure.

In $K^b({\mathsf{F}\mathsf{r}\mathsf{e}\mathsf{e}}_{\mathbb{Z}}^{\mathrm{f}\mathrm{g}})$, let $\mathcal{E}$ be the full subcategory of bounded complexes of finitely generated free abelian groups with even Euler characteristic. Shifts negate Euler characteristic and a cone triangle makes Euler characteristics additive, so $\mathcal{E}$ is triangulated.

The complex $\mathbb{Z}[0]\oplus \mathbb{Z}[0]$ lies in $\mathcal{E}$ but its direct summand $\mathbb{Z}[0]$ does not; hence $\mathcal{E}$ is not thick.

The declared left rotation ends in $-f[1]$, not $f[1]$.

Accordingly the unrolled translate sequence has alternating signed translated arrows; omitting the sign changes the fixed library convention and cannot be used in its cone formulas.