Proof (0MV1) · ScalingStacks

Proof: We first show that $\textnormal{Hom}(S,-)$ is faithful. Again, consider distinguished triangle (5.1):

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

By Lemma 5.3, any object $M\in\mathcal{C}$ is isomorphic to some coproduct $B^{(I)}$ , where $I$ is an indexing set and $B^{(I)}$ denotes the, possibly infinite, coproduct $\coprod_{i\in I}B$ . Hence, to show fidelity we can consider a morphism $B^{(I)}\rightarrow B^{(J)}$ which becomes zero under $\textnormal{Hom}(S,-)$ and show that it is itself necessarily zero. It is sufficient to show that the composite $B\hookrightarrow B^{(I)}\rightarrow B^{(J)}$ is zero, where the morphism $B\hookrightarrow B^{(I)}$ is just the coproduct inclusion into the $i^{\textnormal{th}}$ -summand for each $i\in I$ . This puts us in the following situation:

But, $A\in\Sigma^{-1}\mathcal{A}$ and $B^{(J)}\in\mathcal{C}=\mathcal{A}\cap\mathcal{B}$ , so that $\textnormal{Hom}(A,B^{(J)})=0$ . Therefore, the dotted arrow above is necessarily zero. Hence the composite $B\hookrightarrow B^{(I)}\rightarrow B^{(J)}$ is zero, showing that $\textnormal{Hom}(S,-)$ is faithful.

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

\theta:\textnormal{Hom}(S,B^{(I)})\rightarrow\textnormal{Hom}(S,B^{(J)}),

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

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

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

(5.4)

where $\phi=\textnormal{Hom}(\sigma,B^{(J)})^{-1}\circ\theta\circ\textnormal{Hom}(% \sigma,B^{(I)})$ . Let $q_{i}:B\hookrightarrow B^{(I)}$ be the $i^{\textrm{th}}$ -inclusion of the coproduct and consider its image $\phi(q_{i}):B\rightarrow B^{(J)}$ . By the universal property of the coproduct there exists a unique map $\langle\phi(q_{i})\rangle:B^{(I)}\rightarrow B^{(J)}$ such that the following diagram commutes for each $i\in I$ :

Let us show that $\langle\phi(q_{i})\rangle$ induces $\theta$ under $\textnormal{Hom}(S,-)$ . The map $\textnormal{Hom}(S,\sigma):\textnormal{Hom}(S,S)\rightarrow\textnormal{Hom}(S,B)$ is an isomorphism, therefore, it takes a set of generators for $\textnormal{Hom}(S,S)$ to a set of generators for $\textnormal{Hom}(S,B)$ . The vector space $\textnormal{Hom}(S,S)$ is one-dimensional and generated by the identity map on $S$ , $1_{S}$ , whose image under $\textnormal{Hom}(S,\sigma)$ is $\sigma:S\rightarrow B$ . Hence $\textnormal{Hom}(S,B)$ is generated by $\sigma$ . By the compactness of $S$ , we have

\textnormal{Hom}(S,B^{(I)})\cong\coprod_{I}\textnormal{Hom}(S,B)

and $B^{(I)}$ is generated by $|I|$ copies of $\sigma$ . It follows that $\textnormal{Hom}(S,B^{(I)})$ is generated by the family $\{\sigma\circ q_{i}\}_{i\in I}$ . Therefore, we now only need to check that $\theta$ and the map, $\textnormal{Hom}(S,\langle\phi(q_{i})\rangle)$ , induced by $\langle\phi(q_{i})\rangle$ coincide on this set of generators.

By the commutativity of diagram (5.4) we have:

$\displaystyle\theta(q_{i}\circ\sigma)$	$\displaystyle=$	$\displaystyle\textnormal{Hom}(\sigma,B^{(J)})\circ\phi(q_{i})$
	$\displaystyle=$	$\displaystyle\phi(q_{i})\circ\sigma$
	$\displaystyle=$	$\displaystyle(\langle\phi(q_{i})\rangle\circ q_{i})\circ\sigma$
	$\displaystyle=$	$\displaystyle\langle\phi(q_{i})\rangle\circ(q_{i}\circ\sigma)$
	$\displaystyle=$	$\displaystyle\textnormal{Hom}(S,\langle\phi(q_{i})\rangle)(q_{i}\circ\sigma)$

Hence, $\theta$ and $\textnormal{Hom}(S,\langle\phi(q_{i})\rangle)$ coincide on a basis of $\textnormal{Hom}(S,B^{(I)})$ , thus

\theta=\textnormal{Hom}(S,\langle\phi(q_{i})\rangle)

with $\langle\phi(q_{i})\rangle\in\textnormal{Hom}(B^{(I)},B^{(J)})$ . Therefore, the functor $\textnormal{Hom}(S,-)$ is full and faithful. $\Box$