Proof (0MUH) · ScalingStacks

Proof: Let $M$ be an arbitrary object of $\mathcal{T}$ . We first construct a chain of objects and morphisms,

M=M_{0}\stackrel{{\scriptstyle\mu_{0}}}{{\longrightarrow}}M_{1}\stackrel{{% \scriptstyle\mu_{1}}}{{\longrightarrow}}M_{2}\stackrel{{\scriptstyle\mu_{2}}}{% {\longrightarrow}}M_{3}\stackrel{{\scriptstyle\mu_{3}}}{{\longrightarrow}}% \cdots\stackrel{{\scriptstyle\mu_{n-1}}}{{\longrightarrow}}M_{n}=\bar{M}

with $M_{k}\in S^{\perp_{k}}$ for each $k\geqslant 1$ , inductively using distinguished triangles. Secondly, we verify that the composite of these maps is an $S^{\perp_{n}}$ -preenvelope.

Write $M=M_{0}$ . Let $n=1$ ; we construct an object $M_{1}$ and a morphism $\mu_{0}:M_{0}\rightarrow M_{1}$ such that $\textnormal{Hom}(S,\Sigma M_{1})=0$ . If $\textnormal{Hom}(S,\Sigma M_{0})=0$ then set $M_{1}=M_{0}$ and $\mu_{0}=1_{M_{0}}$ , the identity map on $M_{0}$ . If not, we can choose a, possibly infinite, coproduct $S^{(m_{1})}$ of copies of $S$ and a nonzero morphism $S^{(m_{1})}\rightarrow\Sigma M_{0}$ which becomes a surjection under the functor $\textnormal{Hom}(S,-)$ . Since the endomorphism ring $\textnormal{End}(S)$ is a division ring we can, moreover, choose $m_{1}$ so that this morphism becomes an isomorphism under $\textnormal{Hom}(S,-)$ . We now extend this morphism to a distinguished triangle:

(3.1)

S^{(m_{1})}\rightarrow\Sigma M_{0}\rightarrow\Sigma M_{1}\rightarrow\Sigma S^{% (m_{1})}.

Applying $\textnormal{Hom}(S,-)$ to (3.1) gives the exact sequence:

\textnormal{Hom}(S,S^{(m_{1})})\stackrel{{\scriptstyle\sim}}{{\longrightarrow}% }\textnormal{Hom}(S,\Sigma M_{0})\rightarrow\textnormal{Hom}(S,\Sigma M_{1})% \rightarrow\textnormal{Hom}(S,\Sigma S^{(m_{1})}).

Since $\textnormal{Hom}(S,\Sigma S^{(m_{1})})=0$ , we get $\textnormal{Hom}(S,\Sigma M_{1})=0$ .

Now suppose $k\geqslant 1$ and suppose we have constructed a chain of objects and morphisms

M=M_{0}\stackrel{{\scriptstyle\mu_{0}}}{{\longrightarrow}}M_{1}\stackrel{{% \scriptstyle\mu_{1}}}{{\longrightarrow}}M_{2}\stackrel{{\scriptstyle\mu_{2}}}{% {\longrightarrow}}M_{3}\stackrel{{\scriptstyle\mu_{3}}}{{\longrightarrow}}% \cdots\stackrel{{\scriptstyle\mu_{k-1}}}{{\longrightarrow}}M_{k}

with $M_{i}\in S^{\perp_{i}}$ for $1\leqslant i\leqslant k$ , and where $\mu_{i}:M_{i}\rightarrow M_{i+1}$ is either the identity map or sits in a distinguished triangle

\Sigma^{-(i+1)}S^{(m_{i+1})}\rightarrow M_{i}\stackrel{{\scriptstyle\mu_{i}}}{% {\longrightarrow}}M_{i+1}\rightarrow\Sigma^{-i}S^{(m_{i+1})}.

If $\textnormal{Hom}(S,\Sigma^{k+1}M_{k+1})=0$ then set $M_{k+1}=M_{k}$ and take $\mu_{k}:M_{k}\rightarrow M_{k+1}$ to be the identity map $1_{M_{k}}$ . If not, we can choose a, possibly infinite, coproduct $S^{(m_{k+1})}$ of copies of $S$ and a nonzero morphism $S^{(m_{k+1})}\rightarrow\Sigma^{k+1}M_{k}$ which becomes an isomorphism under $\textnormal{Hom}(S,-)$ , and then extend it to a distinguished triangle:

(3.2)

S^{(m_{k+1})}\rightarrow\Sigma^{k+1}M_{k}\rightarrow\Sigma^{k+1}M_{k+1}% \rightarrow\Sigma S^{(m_{k+1})}.

As above, an argument from the long exact sequence of Hom-sets arising from (3.2) shows that

\textnormal{Hom}(S,\Sigma^{i}M_{k+1})=0\textnormal{ for }i=1,\ldots,k+1.

The case $i=k$ follows by the injectivity of $\textnormal{Hom}(S,S^{(m_{k+1})})\stackrel{{\scriptstyle\sim}}{{% \longrightarrow}}\textnormal{Hom}(S,\Sigma^{k+1}M_{k})$ ; and the case $i=k+1$ by its surjectivity.

Hence, inductively we obtain a chain of objects and morphisms of $\mathcal{T}$ ,

(3.3)

M=M_{0}\stackrel{{\scriptstyle\mu_{0}}}{{\longrightarrow}}M_{1}\stackrel{{% \scriptstyle\mu_{1}}}{{\longrightarrow}}M_{2}\stackrel{{\scriptstyle\mu_{2}}}{% {\longrightarrow}}M_{3}\stackrel{{\scriptstyle\mu_{3}}}{{\longrightarrow}}% \cdots\stackrel{{\scriptstyle\mu_{n-1}}}{{\longrightarrow}}M_{n},

where each map $\mu_{k}:M_{k}\rightarrow M_{k+1}$ is either the identity map or sits in a distinguished triangle

(3.4)

\Sigma^{-(k+1)}S^{(m_{k+1})}\rightarrow M_{k}\stackrel{{\scriptstyle\mu_{k}}}{% {\longrightarrow}}M_{k+1}\rightarrow\Sigma^{-k}S^{(m_{k+1})}.

To see that the composite $\mu=\mu_{n-1}\circ\cdots\circ\mu_{1}\circ\mu_{0}$ from (3.3) is an $S^{\perp_{n}}$ -preenvelope, we shall show that for each $X\in S^{\perp_{n}}$ the map $\textnormal{Hom}(M_{k+1},X)\rightarrow\textnormal{Hom}(M_{k},X)$ induced by $\mu_{k}$ is a surjection. Without loss of generality we may assume that each map $\mu_{k}$ sits in a distinguished triangle (3.4) above, because if $\mu_{k}=1_{M_{k}}$ , then the map $\textnormal{Hom}(M_{k+1},X)\rightarrow\textnormal{Hom}(M_{k},X)$ is trivially an isomorphism for all $X\in\mathcal{T}$ .

Let $X\in S^{\perp_{n}}$ ; applying $\textnormal{Hom}(-,X)$ to distinguished triangle (3.4), we get the long exact sequence of Hom-sets below:

(\Sigma^{-k}S^{(m_{k+1})},X)\rightarrow(M_{k+1},X)\rightarrow(M_{k},X)% \rightarrow(\Sigma^{-(k+1)}S^{(m_{k+1})},X),

where we have written $(A,B)$ as a shorthand for $\textnormal{Hom}(A,B)$ . Now since we have $\textnormal{Hom}(S^{(m_{k+1})},\Sigma^{k}X)=\textnormal{Hom}(S^{(m_{k+1})},% \Sigma^{(k+1)}X)=0$ for $k=1,\ldots,n-1$ , the map

\textnormal{Hom}(M_{k+1},X)\rightarrow\textnormal{Hom}(M_{k},X)

induced by $\mu_{k}$ is an isomorphism for $k=1,\ldots,n-1$ and a surjection for $k=0$ . Hence, writing $\bar{M}=M_{n}$ , the composite $\mu:M\rightarrow\bar{M}$ is an $S^{\perp_{n}}$ -preenvelope. $\Box$