Lemma 3.20 (0NX7) · ScalingStacks

0NX7

Lemma 3.20. Suppose $p:X\to Y$ is a perfect morphism over a perfect base. Then compact and dualizable objects of $\qc(X)$ coincide.

0NX8

Proof. First note that pushforward $p_{*}$ along the perfect morphism $p$ is colimit preserving, hence (by adjunction) the pullback $p^{*}M$ of a compact object $M\in\qc(Y)$ is compact. In particular we find that the structure sheaf (the monoidal unit) on $X$ is compact, and hence that all dualizable objects are compact. Note also that the pullback of a dualizable object is always dualizable.

Now suppose that $U\to Y$ is any affine mapping to $Y$ , and consider the base change $X_{U}$ of $X$ to $U$ . By the definition of a perfect morphism applied to $p$ , this base change is itself a perfect stack. Let $q:X_{U}\to X$ denote the base change morphism, which is affine since $Y$ has affine diagonal. If $M\in\qc(X)$ is any compact object and $M_{U}=q^{*}M$ is its pullback to $U$ , then $M_{U}$ is itself compact since $q_{*}$ preserves colimits:

$\displaystyle\Hom(M_{U},\colim A_{i})$	$\displaystyle\simeq$	$\displaystyle\Hom(q^{*}M,\colim A_{i})$
	$\displaystyle\simeq$	$\displaystyle\Hom(M,q_{*}\colim A_{i})$
	$\displaystyle\simeq$	$\displaystyle\Hom(M,\colim q_{*}A_{i})$
	$\displaystyle\simeq$	$\displaystyle\colim\Hom(M,q_{*}A_{i})$
	$\displaystyle\simeq$	$\displaystyle\colim\Hom(M_{U},A_{i}).$

Since $U$ itself is perfect, it follows that $M_{U}$ is dualizable and (by Proposition 3.6) perfect. We now show that the pullback of $M$ to any affine is perfect. Assume that the map $U\rightarrow Y$ is surjective, so as a consequence $X_{U}\rightarrow X$ is also surjective. Now let $f:V\rightarrow X$ be any affine mapping to $X$ . We may form the Cartesian diagrams:

We now verify that $f^{*}M$ is perfect, given the hypotheses above. The fiber product $U\times_{Y}V$ is affine, since $Y$ has affine diagonal, and the map $q^{\prime}$ is surjective since $q$ is. Since $M_{U}$ is perfect, $f^{\prime*}M_{U}\simeq q^{\prime*}f^{*}M$ is perfect, as well. Thus, in summary, we know that $q^{\prime*}f^{*}M=\mathcal{O}(V\times_{Y}U)\otimes_{\mathcal{O}(U)}f^{*}M$ is perfect, where $q^{\prime}$ is surjective, and hence $f^{*}M$ is perfect. Since the pullback of $M$ to any affine is perfect, therefore $M$ is itself perfect by Definition 3.1 and hence (again by Proposition 3.6) dualizable. ∎