ScalingStacks

Hence, inductively we obtain a chain of objects and morphisms of $𝒯$,

(3.3)

$$M=M_0\stackrel{{\mu }_0}{⟶}{M}_1\stackrel{{\mu }_1}{⟶}{M}_2\stackrel{{\mu }_2}{⟶}{M}_3\stackrel{{\mu }_3}{⟶}\mathrm{\cdots }\stackrel{{\mu }_{n-1}}{⟶}{M}_n,$$

where each map ${\mu }_k:M_k\to M_{k+1}$ is either the identity map or sits in a distinguished triangle

(3.4)

$${\mathrm{\Sigma }}^{-(k+1)}{S}^{(m_{k+1})}\to M_k\stackrel{{\mu }_k}{⟶}{M}_{k+1}\to {\mathrm{\Sigma }}^{-k}{S}^{(m_{k+1})}.$$

To see that the composite $\mu ={\mu }_{n-1}\circ \mathrm{\cdots }\circ {\mu }_1\circ {\mu }_0$ from (3.3) is an $S^{{⟂}_n}$-preenvelope, we shall show that for each $X\in S^{{⟂}_n}$ the map $\text{Hom}(M_{k+1},X)\to \text{Hom}(M_k,X)$ 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 $\text{Hom}(M_{k+1},X)\to \text{Hom}(M_k,X)$ is trivially an isomorphism for all $X\in 𝒯$.