Reference (0MVN) · ScalingStacks

Proof: Consider the object $S$ of $\mathcal{T}$ . Since $(\mathcal{A},\mathcal{B})$ forms a co- $t$ -structure on $\mathcal{T}$ , there is a distinguished triangle

(5.1)

\Sigma^{-1}A\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}S\longrightarrow B% \longrightarrow A{}{}

with $A\in\mathcal{A}$ and $B\in\mathcal{B}$ . Applying the functor $\textnormal{Hom}(S,-)$ to (5.1) gives the following long exact sequence:

(5.2)

\displaystyle\textnormal{Hom}(S,\Sigma^{i}S)\rightarrow\textnormal{Hom}(S,% \Sigma^{i}B)\rightarrow\textnormal{Hom}(S,\Sigma^{i}A).

In (5.2) we have $\textnormal{Hom}(S,\Sigma^{i}S)=\textnormal{Hom}(S,\Sigma^{i}A)=0$ for all $i<0$ since $A\in\mathcal{A}$ , so that $\textnormal{Hom}(S,\Sigma^{i}B)=0$ for all $i<0$ . We know that $\textnormal{Hom}(S,\Sigma^{i}B)=0$ for all $i>0$ since $B\in\mathcal{B}$ . Therefore, $\textnormal{Hom}(S,\Sigma^{i}B)=0$ for all $i\neq 0$ . By Lemma 4.3, $\textnormal{Hom}(S,S)\rightarrow\textnormal{Hom}(S,B)$ is an isomorphism. Hence we have:

\textnormal{Hom}(S,\Sigma^{i}B^{(m)})=\left\{\begin{array}[]{ll}0&\textrm{if $% i\neq 0$}\\ k^{(m)}&\textrm{if $i=0$.}\end{array}\right.

where $k=\textnormal{End}(S)$ . Hence $\textnormal{Hom}(S,-):\mathcal{C}\rightarrow\textnormal{Mod}(k^{\textnormal{op% }})$ is dense. $\Box$

David Pauksztello

Original source: arXiv:0705.0102v2