Lemma 22.27.15. In Situation 22.27.2 given admissible monomorphisms x \xrightarrow {\alpha } y, y \xrightarrow {\beta } z in \mathcal{A}, there exist distinguished triangles (x,y,q_1,\alpha ,p_1,\delta _1), (x,z,q_2,\beta \alpha ,p_2,\delta _2) and (y,z,q_3,\beta ,p_3,\delta _3) for which TR4 holds.
Proof. Given admissible monomorphisms x\xrightarrow {\alpha } y and y\xrightarrow {\beta }z, we can find distinguished triangles, via their extensions to admissible short exact sequences,
\xymatrix{ x\ar@<0.5ex>[r]^{\alpha } & y\ar@<0.5ex>[l]^{\pi _1} \ar@<0.5ex>[r]^{p_1} & q_1 \ar[r]^{\delta _1} \ar@<0.5ex>[l]^{s_1} & x[1] }
\xymatrix{ x\ar@<0.5ex>[r]^{\beta \alpha } & z\ar@<0.5ex>[l]^{\pi _1\pi _3} \ar@<0.5ex>[r]^{p_2} & q_2 \ar[r]^{\delta _2} \ar@<0.5ex>[l]^{s_2} & x[1] }
\xymatrix{ y\ar@<0.5ex>[r]^{\beta } & z\ar@<0.5ex>[l]^{\pi _3} \ar@<0.5ex>[r]^{p_3} & q_3 \ar[r]^{\delta _3} \ar@<0.5ex>[l]^{s_3} & x[1] }
In these diagrams, the maps \delta _ i are defined as \delta _ i = \pi _ i d(s_ i) analogous to the maps defined in Lemma 22.27.1. They fit in the following solid commutative diagram
\xymatrix@C=5pc@R=3pc{ x\ar@<0.5ex>[r]^{\alpha } \ar@<0.5ex>[dr]^{\beta \alpha } & y\ar@<0.5ex>[d]^{\beta } \ar@<0.5ex>[l]^{\pi _1} \ar@<0.5ex>[r]^{p_1} & q_1 \ar[r]^{\delta _1} \ar@<0.5ex>[l]^{s_1} \ar@{.>}[dd]^{p_2\beta s_1} & x[1] \\ & z \ar@<0.5ex>[u]^{\pi _3}\ar@<0.5ex>[d]^{p_3} \ar@<0.5ex>[dr]^{p_2} \ar@<0.5ex>[ul]^{\pi _1\pi _3} & & \\ & q_3\ar@<0.5ex>[u]^{s_3} \ar[d]^{\delta _3} & q_2 \ar@{.>}[l]^{p_3s_2} \ar@<0.5ex>[ul]^{s_2} \ar[dr]^{\delta _2} \\ & y[1] & & x[1]}
where we have defined the dashed arrows as indicated. Clearly, their composition p_3s_2p_2\beta s_1 = 0 since s_2p_2 = 0. We claim that they both are morphisms of \text{Comp}(\mathcal{A}). We can check this using equations in Lemma 22.27.1:
d(p_2\beta s_1) = p_2\beta d(s_1) = p_2\beta \alpha \pi _1 d(s_1) = 0
since p_2\beta \alpha = 0, and
d(p_3s_2) = p_3d(s_2) = p_3\beta \alpha \pi _1\pi _3 d(s_2) = 0
since p_3\beta = 0. To check that q_1\to q_2\to q_3 is an admissible short exact sequence, it remains to show that in the underlying graded category, q_2 = q_1\oplus q_3 with the above two morphisms as coprojection and projection. To do this, observe that in the underlying graded category \mathcal{C}, there hold
y = x\oplus q_1,\quad z = y\oplus q_3 = x\oplus q_1\oplus q_3
where \pi _1\pi _3 gives the projection morphism onto the first factor: x\oplus q_1\oplus q_3\to z. By axiom (A) on \mathcal{A}, \mathcal{C} is an additive category, hence we may apply Homology, Lemma 12.3.10 and conclude that
\mathop{\mathrm{Ker}}(\pi _1\pi _3) = q_1\oplus q_3
in \mathcal{C}. Another application of Homology, Lemma 12.3.10 to z = x\oplus q_2 gives \mathop{\mathrm{Ker}}(\pi _1\pi _3) = q_2. Hence q_2\cong q_1\oplus q_3 in \mathcal{C}. It is clear that the dashed morphisms defined above give coprojection and projection.
Finally, we have to check that the morphism \delta : q_3 \to q_1[1] induced by the admissible short exact sequence q_1\to q_2\to q_3 agrees with p_1\delta _3. By the construction in Lemma 22.27.1, the morphism \delta is given by
\begin{align*} p_1\pi _3s_2d(p_2s_3) = & p_1\pi _3s_2p_2d(s_3) \\ = & p_1\pi _3(1-\beta \alpha \pi _1\pi _3)d(s_3) \\ = & p_1\pi _3d(s_3)\quad (\text{since }\pi _3\beta = 0) \\ = & p_1\delta _3 \end{align*}
as desired. The proof is complete. \square
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