Lemma 5.3 (0MUY) · ScalingStacks

0MUY

Lemma 5.3. Under the hypotheses of Setup 5.1 we have that each $M\in\mathcal{C}$ is isomorphic to $B^{(m)}$ for some $m$ .

0MUZ

Proof: Consider the distinguished triangle (5.1) from Lemma 5.2:

\displaystyle\Sigma^{-1}A\rightarrow S\rightarrow B\rightarrow A

with $A\in\Sigma^{-1}\mathcal{A}$ and $B\in\mathcal{C}$ . Let $M\in\mathcal{C}$ ; since $k=\textnormal{Hom}(S,S)$ is a skew field, we can choose $m$ such that $S^{(m)}\rightarrow M$ becomes an isomorphism under $\textnormal{Hom}(S,-)$ . Again, by Lemma 4.3, the morphism $S\rightarrow B$ becomes an isomorphism under $\textnormal{Hom}(S,-)$ .

We may now apply the functor $\textnormal{Hom}(-,M)$ to (5.1) and from the long exact sequence notice that the morphism $\textnormal{Hom}(B,M)\rightarrow\textnormal{Hom}(S,M)$ is an isomorphism. So we obtain the commutative diagram:

where both $S^{(m)}\rightarrow M$ and $S^{(m)}\rightarrow B^{(m)}$ are isomorphisms under $\textnormal{Hom}(S,-)$ . Hence the unique map $B^{(m)}\rightarrow M$ making the diagram above commute becomes an isomorphism under $\textnormal{Hom}(S,-)$ .

Now extend this unique map $B^{(m)}\rightarrow M$ to a distinguished triangle

(5.3)

\displaystyle B^{(m)}\rightarrow M\rightarrow Z\rightarrow\Sigma B^{(m)}

and apply the functor $\textnormal{Hom}(S,-)$ to give a long exact sequence. One easily sees from this long exact sequence that $\textnormal{Hom}(S,\Sigma^{i}Z)=0$ for $i\neq 0$ . The fact that the morphism $B^{(m)}\rightarrow M$ becomes the isomorphism $\textnormal{Hom}(S,B^{(m)})\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}% \textnormal{Hom}(S,M)$ forces $\textnormal{Hom}(S,Z)=0$ so that $\textnormal{Hom}(S,\Sigma^{i}Z)=0$ for all $i\in\mathbb{Z}$ . Since $\{\Sigma^{i}S\ |\ i\in\mathbb{Z}\}$ is a generating set for $\mathcal{T}$ , it follows that $Z=0$ . Hence $B^{(m)}\rightarrow M$ is an isomorphism. $\Box$