ScalingStacks

Proof. The second assertion follows from the first and Proposition 3.21, since $X_{1}\times_{Y}X_{2}\to X_{1}\times X_{2}$ is an affine morphism for $Y$ with affine diagonal.

To prove the first assertion, note that $X=X_{1}\times X_{2}$ has affine diagonal. Applying Lemma 3.20 to the projection to a factor, we see that compact and dualizable objects of $\qc(X)$ coincide. Thus to confirm that $X$ is perfect, it suffices to show that $\qc(X)$ is compactly generated.

Let us first check that the external product of compact objects is again compact. By assumption, compact objects $M_{i}$ on the factors $X_{i}$ are dualizable, so $M_{1}\boxtimes M_{2}$ is dualizable (it is the tensor product of pullbacks, and both operations preserve dualizabie objects), and hence compact.

Now let us check that external products of compact objects generate $\qc(X)$. The argument is a modification of the argument of Bondal-Van den Bergh [BV] in the case of a single compact generator. Namely, let $N$ be right orthogonal to $\qc(X)^{c}$, so that in particular $\Hom(M_{1}\boxtimes M_{2},N)\simeq 0$ for all $M_{i}\in\qc(X_{i})^{c}$. By adjunction, we have

	$\displaystyle 0$	$\displaystyle\simeq$	$\displaystyle\Hom(M_{1}\boxtimes M_{2},N)$
		$\displaystyle\simeq$	$\displaystyle\Hom(\pi_{1}^{*}M_{1},{\mathcal{H}om}(\pi_{2}^{*}M_{2},N))$
		$\displaystyle\simeq$	$\displaystyle\Hom(M_{1},\pi_{1,*}{\mathcal{H}om}(\pi_{2}^{*}M_{2},N))$

for all $M_{1},M_{2}$, so that $\pi_{1*}{\mathcal{H}om}(\pi_{2}^{*}M_{2},N)\simeq 0$ since such $M_{1}$ generate $\qc(X_{1})$. For any affines $U\to X_{1}$ and $V\to X_{2}$, we therefore have

	$\displaystyle 0$	$\displaystyle\simeq$	$\displaystyle\Gamma(U,\pi_{1,*}{\mathcal{H}om}(\pi_{2}^{*}M_{2},N))$
		$\displaystyle\simeq$	$\displaystyle\Hom_{U\times X_{2}}(\pi_{2}^{*}M_{2},N)$
		$\displaystyle\simeq$	$\displaystyle\Hom_{X_{2}}(M_{2},(\pi_{2}|_{U\times X_{2}})_{*}N)$

for all $M_{2}$. Since the latter objects generate $\qc(X_{2})$ it follows (upon restricting to $V$) that $\Gamma(U\times V,N)\simeq 0$, whence (by affineness of $U\times V$) that $N|_{U\times V}\simeq 0$, and finally (since affines of the form $U\times V$ cover $X_{1}\times X_{2}$) that $N\simeq 0$. ∎