ScalingStacks

We must also show that $\text{Hom}(S,-)$ is full. Suppose we have a morphism

$$\theta :\text{Hom}(S,B^{(I)})\to \text{Hom}(S,B^{(J)}),$$

where $I$ and $J$ are again indexing sets. We must construct a morphism $B^{(I)}\to B^{(J)}$ which induces $\theta$ under $\text{Hom}(S,-)$. We recall distinguished triangle (5.1) again:

$${\mathrm{\Sigma }}^{-1}A⟶S\stackrel{\sigma }{⟶}B⟶A.$$

Note that $\sigma :S\to B$ becomes an isomorphism under $\text{Hom}(-,B^{(I)})$ because $B^{(I)}\in 𝒞$. Hence we get the following commutative diagram:

(5.4)

where $\varphi =\text{Hom}{(\sigma ,B^{(J)})}^{-1}\circ \theta \circ \text{Hom}(\sigma ,B^{(I)})$. Let $q_i:B\hookrightarrow B^{(I)}$ be the $i^{\text{th}}$-inclusion of the coproduct and consider its image $\varphi (q_i):B\to B^{(J)}$. By the universal property of the coproduct there exists a unique map $⟨\varphi (q_i)⟩:B^{(I)}\to B^{(J)}$ such that the following diagram commutes for each $i\in I$: