Proof. Proposition 9.20 implies that V ( − ) V(-) inverts Morita
equivalences. Furthermore, by Lemma 9.7 , ℱ κ {\mathcal{F}}_{\kappa} and Σ κ \Sigma_{\kappa} preserve
κ \kappa -filtered colimits for κ > ω \kappa>\omega ,
and so V ( − ) V(-) does as well. Now, let
𝒜 ⟶ ℬ ⟶ 𝒞 {\mathcal{A}}\longrightarrow{\mathcal{B}}\longrightarrow{\mathcal{C}}
be an exact sequence. Proposition 9.20 implies that we
can assume that Ho ( 𝒜 ) \Ho({\mathcal{A}}) is a thick triangulated subcategory of
Ho ( ℬ ) \Ho({\mathcal{B}}) . Consider the following diagram
(9.22)
Dia ( 𝒜 ) \textstyle{\mathrm{Dia}({\mathcal{A}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces} Dia ( ℬ ) \textstyle{\mathrm{Dia}({\mathcal{B}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces} Dia ( 𝒜 , ℬ ) := Dia ( 𝒜 ) / Dia ( ℬ ) \textstyle{\mathrm{Dia}({\mathcal{A}},{\mathcal{B}}):=\mathrm{Dia}({\mathcal{A}})/\mathrm{Dia}({\mathcal{B}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces} D \scriptstyle{D} Dia ( 𝒜 ) \textstyle{\mathrm{Dia}({\mathcal{A}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces} Dia ( ℬ ) \textstyle{\mathrm{Dia}({\mathcal{B}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces} Dia ( 𝒞 ) , \textstyle{\mathrm{Dia}({\mathcal{C}})\,,}
where Dia ( 𝒜 , ℬ ) \mathrm{Dia}({\mathcal{A}},{\mathcal{B}}) is obtained by passage to the cofiber
objectwise. Note that since in the above diagram (9.22 )
the upper row is objectwise a strict-exact sequence, we obtain a
cofiber sequence
V ( 𝒜 ) ⟶ V ( ℬ ) ⟶ V ( ℬ , 𝒜 ) ⟶ Σ V ( 𝒜 ) V({\mathcal{A}})\longrightarrow V({\mathcal{B}})\longrightarrow V({\mathcal{B}},{\mathcal{A}})\longrightarrow\Sigma V({\mathcal{A}})
in ℳ wloc κ ¯ \underline{{\mathcal{M}}_{\mathrm{wloc}}^{\kappa}} , where
V ( ℬ , 𝒜 ) := colim n Σ − n 𝒰 wloc κ ¯ ( Σ κ ( n ) ( ℬ ) / Σ κ ( n ) ( 𝒜 ) ) . V({\mathcal{B}},{\mathcal{A}}):=\colim_{n}\Sigma^{-n}\underline{{\mathcal{U}}_{\mathrm{wloc}}^{\kappa}}(\Sigma_{\kappa}^{(n)}({\mathcal{B}})/\Sigma_{\kappa}^{(n)}({\mathcal{A}}))\,.
We now show that the induced map
(9.23)
V ( ℬ , 𝒜 ) ⟶ V ( 𝒞 ) V({\mathcal{B}},{\mathcal{A}})\longrightarrow V({\mathcal{C}})
is an equivalence. For this,
consider the following commutative diagram
Σ κ ( n ) ( 𝒜 ) \textstyle{\Sigma_{\kappa}^{(n)}({\mathcal{A}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces} ℱ κ Σ κ ( n ) ( 𝒜 ) \textstyle{{\mathcal{F}}_{\kappa}\Sigma_{\kappa}^{(n)}({\mathcal{A}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces} Σ κ ( n + 1 ) ( 𝒜 ) \textstyle{\Sigma_{\kappa}^{(n+1)}({\mathcal{A}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces} Σ κ ( n ) ( ℬ ) \textstyle{\Sigma_{\kappa}^{(n)}({\mathcal{B}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces} ℱ κ Σ κ ( n ) ( ℬ ) \textstyle{{\mathcal{F}}_{\kappa}\Sigma_{\kappa}^{(n)}({\mathcal{B}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces} Σ κ ( n + 1 ) ( ℬ ) \textstyle{\Sigma_{\kappa}^{(n+1)}({\mathcal{B}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces} Σ κ ( n ) ( ℬ ) / Σ κ ( n ) ( 𝒜 ) \textstyle{\Sigma_{\kappa}^{(n)}({\mathcal{B}})/\Sigma_{\kappa}^{(n)}({\mathcal{A}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces} ℱ κ Σ κ ( n ) ( ℬ ) / ℱ κ Σ κ ( n ) ( 𝒜 ) \textstyle{{\mathcal{F}}_{\kappa}\Sigma_{\kappa}^{(n)}({\mathcal{B}})/{\mathcal{F}}_{\kappa}\Sigma_{\kappa}^{(n)}({\mathcal{A}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces} θ n \scriptstyle{\theta_{n}} Σ κ ( n + 1 ) ( ℬ ) / Σ κ ( n + 1 ) ( 𝒜 ) \textstyle{\Sigma_{\kappa}^{(n+1)}({\mathcal{B}})/\Sigma_{\kappa}^{(n+1)}({\mathcal{A}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces} D n \scriptstyle{D_{n}} Σ κ ( n ) ( 𝒞 ) \textstyle{\Sigma_{\kappa}^{(n)}({\mathcal{C}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces} ℱ κ Σ κ ( n ) ( 𝒞 ) \textstyle{{\mathcal{F}}_{\kappa}\Sigma_{\kappa}^{(n)}({\mathcal{C}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces} Σ κ ( n + 1 ) ( 𝒞 ) . \textstyle{\Sigma_{\kappa}^{(n+1)}({\mathcal{C}})\,.}
Since the induced triangulated functor
Ho ( ℱ κ Σ κ ( n ) ( 𝒜 ) ) ⟶ Ho ( ℱ κ Σ κ ( n ) ( ℬ ) ) \Ho({\mathcal{F}}_{\kappa}\Sigma_{\kappa}^{(n)}({\mathcal{A}}))\longrightarrow\Ho({\mathcal{F}}_{\kappa}\Sigma_{\kappa}^{(n)}({\mathcal{B}}))
preserves κ \kappa -small colimits, [70 , §3.1] implies that
the triangulated category
Ho ( ℱ κ Σ κ ( n ) ( ℬ ) / ℱ κ Σ κ ( n ) ( 𝒜 ) ) \Ho({\mathcal{F}}_{\kappa}\Sigma_{\kappa}^{(n)}({\mathcal{B}})/{\mathcal{F}}_{\kappa}\Sigma_{\kappa}^{(n)}({\mathcal{A}}))
is idempotent complete. Therefore, θ n \theta_{n} is an
equivalence, and we obtain maps
ψ n : Σ κ ( n ) ( 𝒞 ) ⟶ ℱ κ Σ κ ( n ) ( ℬ ) / ℱ κ Σ κ ( n ) ( 𝒜 ) \psi_{n}:\Sigma_{\kappa}^{(n)}({\mathcal{C}})\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}{\mathcal{F}}_{\kappa}\Sigma_{\kappa}^{(n)}({\mathcal{B}})/{\mathcal{F}}_{\kappa}\Sigma_{\kappa}^{(n)}({\mathcal{A}})
which induce maps
Ψ n : Σ − n 𝒰 wloc κ ¯ ( Σ κ ( n ) ( 𝒞 ) ) ⟶ Σ − n − 1 𝒰 wloc κ ¯ ( Σ κ ( n + 1 ) ( ℬ ) / Σ κ ( n + 1 ) ( 𝒜 ) ) . \Psi_{n}:\Sigma^{-n}\underline{{\mathcal{U}}_{\mathrm{wloc}}^{\kappa}}(\Sigma_{\kappa}^{(n)}({\mathcal{C}}))\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\Sigma^{-n-1}\underline{{\mathcal{U}}_{\mathrm{wloc}}^{\kappa}}(\Sigma_{\kappa}^{(n+1)}({\mathcal{B}})/\Sigma^{(n+1)}_{\kappa}({\mathcal{A}})).
It follows that the natural map
colim n Σ − n 𝒰 wloc κ ¯ ( Σ κ ( n ) ( ℬ ) / Σ κ ( n ) ( 𝒜 ) ) ⟶ colim n Σ − n 𝒰 wloc κ ¯ ( Σ κ ( n ) ( 𝒞 ) ) \colim_{n}\Sigma^{-n}\underline{{\mathcal{U}}_{\mathrm{wloc}}^{\kappa}}(\Sigma_{\kappa}^{(n)}({\mathcal{B}})/\Sigma_{\kappa}^{(n)}({\mathcal{A}}))\mathchoice{\longrightarrow}{\rightarrow}{\rightarrow}{\rightarrow}\colim_{n}\Sigma^{-n}\underline{{\mathcal{U}}_{\mathrm{wloc}}^{\kappa}}(\Sigma_{\kappa}^{(n)}({\mathcal{C}}))
is an equivalence, which implies that the map (9.23 ) is an equivalence.
∎