Opposite and product of derived categories Let $\mathscr{A, B}$ be abelian categories.
Then $D(\mathscr{A}) \times D(\mathscr{B}) = D(\mathscr{A \times B})$?
And $D(\mathscr{A})^{\text{op}} = D(\mathscr{A}^{\text{op}})$?
I can't find any references mentioning it.
Please check my "proof".
First, by the definition of product of categories, these equation holds for the category of complexes.
Next, since "being homotopic equivalent" is preserved by product and opposite, these equations hold for the homotopy category.
For the opposite category:
Let $K(\mathscr{A})^{\text{op}} \to \mathscr{C}$ be a $\partial$-functor of triangulated categories which takes quasi-isomorphisms to isomorphisms.
Then the opposite $K(\mathscr{A}) \to \mathscr{C}^{\text{op}}$ has the same propeties.
So this factors uniquely through $K(\mathscr{A}) \to D(\mathscr{A})$.
Thus $D(\mathscr{A})^{\text{op}}$ is the localization of $K(\mathscr{A})^{\text{op}}$.
So using $K(\mathscr{A})^{\text{op}} = K(\mathscr{A}^{\text{op}})$, we have $D(\mathscr{A})^{\text{op}} \cong D(\mathscr{A}^{\text{op}})$ by a unique isomorphism.
For the product:
Let $K(\mathscr{A}) \times K(\mathscr{B}) = K(\mathscr{A \times B}) \to \mathscr{C}$ be a $\partial$-functor which takes quasi-isomophisms to isomorphisms.
Then by the same argument, using $K(\mathscr{A}) \to K(\mathscr{A}) \times K(\mathscr{B}): A \mapsto (A, 0)$, we have two induced maps $D(\mathscr{A}) \to \mathscr{C}$ and $D(\mathscr{B}) \to \mathscr{C}$.
So by the "universal property" (2-universal property?) of product of category, we have $D(\mathscr{A}) \times D(\mathscr{B}) \to \mathscr{C}$.
Thus $K(\mathscr{A}) \times K(\mathscr{B}) \to D(\mathscr{A}) \times D(\mathscr{B})$ is the localization.
So $D(\mathscr{A}) \times D(\mathscr{B}) \cong D(\mathscr{A} \times \mathscr{B})$ by a unique isomorphism.
I want these equation to show that $- \otimes^{\mathbb{L}^-}- = \mathbb{L}^-(- \otimes -)$ and $\mathbb{R}^+\operatorname{Hom}(-,-) = \mathbb{R}^+(\operatorname{Hom}(-,-))$ for the category of sheaves.
(Where the left sides are defined by the unique functor that induced by a functor $K \times D \to D$, and the right sides are derived functors of two variable tensor functor (resp. two variable hom. functor) $ K(\mathscr{A}) \times K(\mathscr{B}) = K(\mathscr{A \times B}) \to \mathscr{A}b$.)
Thank you very much!
 A: The product of categories is not a coproduct, so your proof for products doesn't quite work. You could come up with a proof using the universal property, but you should use cartesian closedness of the category of categories: functors $K(\mathcal A)\times K(\mathcal B)\to \mathcal C$ correspond to functors $K(\mathcal A)\to \mathcal C^{K(\mathcal B)}$. 
Consider the class of quasi-isomorphisms of $K(\mathcal A)\times K(\mathcal B)$, a category whose objects are pairs $(A_\bullet,B_\bullet)$. The homology of such a pair is $(H_n(A_\bullet),H_n(B_\bullet))$. Thus the quasi-isomorphisms in $K(\mathcal A)\times K(\mathcal B)$ are precisely the maps whose projections to $K(\mathcal A)$ and to $K(\mathcal B)$ are quasi-isomorphisms. 
In other words, it suffices to show, for categories $C$ and $D$ with classes $W_C$ and $W_D$ of morphisms, that $C\times D[(W_C\times W_D)^{-1}]\simeq C[W_C^{-1}]\times D[W_D^{-1}]$. This isn't quite true in general, but it suffices to assume that $W_C$ and $W_D$ contain the identities. 
Now, a functor $$F_1:C[W_C^{-1}]\times D[W_D^{-1}]\to E$$ is equivalently a functor $$F_2:C[W_C^{-1}]\to E^{D[W_D^{-1}]},$$ which is equivalently
a functor $$F_3:C\to E^{D[W_D^{-1}]}$$ inverting $W_C$. Thus if $f\in W_C$ then $F_3(f)$ is a natural isomorphism-its components are isomorphisms in $E$.
This shows that $F_3$ is equivalently a functor $$F_4:D[W_D^{-1}]\to E^C$$ factoring through the full subcategory on objects of $E^C$ inverting $W_C$, which is equivalently a functor $$F_5:D\to E^C$$ inverting $W_D$ and satisfying the same factorization as $F_4$. But this is equivalently a functor $$F_6:C\times D\to E$$ inverting morphisms of the form $(1_c,g)$ or $(f,1_d)$ where $f\in W_C$ and $g\in W_D$. Finally, this condition is equivalent to $F_6$ inverting $W_C\times W_D$, since $W_C$ and $W_D$ contain the identities. 
This has been a somewhat unpleasant but very formal exercise, so surely there's a more abstract approach. And indeed there is! The 2-category, not just the category, of categories is cartesian closed. This means that the product 2-functor $(-)\times A$ is a left 2-adjoint, not just a left adjoint, which is just a fancy way of saying that the category of functors $B\times A\to C$ is isomorphic to the category of functors $B\to C^A$, and that's better than a bijection of sets. 
Now, as a left 2-adjoint, $(-)\times A$ preserves all 2-categorical colimits. In particular, it preserves coinverters, which are functors universally making a given natural transformation invertible. Localizations are examples of coinverters, so we can write $$C[W_C^{-1}]\times D[W_D^{-1}]\cong (C\times D[W_D^{-1}])[(W_C\times I_D)^{-1}]$$$$\cong (C\times D)[(I_C\times W_D)^{-1}][(W_C\times I_D)^{-1}]\cong C\times D[(W_C\times W_D)^{-1}]$$
where $I_C$ and $I_D$ denote the respective classes of isomorphisms, and the last step again used that $W_C$ and $W_D$ contain $I_C$ and $I_D$, respectively, while the first two steps use preservation of coinverters by products. 
Incidentally, a terminological complaint: $D(\mathcal A)$ is just the localization of $K(\mathcal A)$ at the quasi-isomorphisms. There's no need to talk about $\delta$-functors once we've passed from $\mathcal A$ to $K(\mathcal A)$. 
