Let $X$ and $Y$ be two quasi compact, separated schemes over $k$, and consider the fibre product $X \times Y$. If we call $p_1$ and $p_2$ the two projections, and we take perfect complexes $F_1, F_2 \in \mathfrak{Perf}(X)$, $G_1, G_2 \in \mathfrak{Perf}(Y)$, then by flat base change we have an isomorphism $$ \text{RHom}_{X}(p_1^{\ast}F_1, p_1^{\ast}F_2) \otimes \text{RHom}_{Y}(p_2^{\ast}G_1, p_2^{\ast}G_2) \simeq \text{RHom}_{X \times Y}(p_1^{\ast}F_1 \otimes p_2^{\ast}G_1, p_1^{\ast}F_2 \otimes p_2^{\ast}G_2). $$ I am trying to understand whether this is true even when we take all the complexes to be in $D_{qc}(X)$ and $D_{qc}(Y)$.
I am fairly positive, because by a result of Ben-Zvi, Francis, Nadler - arXiv:0805.0157 - we know that if consider DG enhancements of these categories (which I will denote by italic letters), then there is a quasi equivalence $$ \mathcal{D}_X \otimes \mathcal{D}_Y \simeq \mathcal{D}_{X \times Y}. $$ However, while the right hand side has homotopy category $D_{qc}(X \times Y)$, the tensor product on the left hand side might mess things up at the level of homotopy categories. My idea was to try to apply the indization functor to the fully faithful functor $$ \mathfrak{Perf}(X) \otimes \mathfrak{Perf}(Y) \rightarrow \mathfrak{Perf}(X \times Y), $$ but the indization functor is symmetric monodical only when we consider the tensor product of stale $\infty$ categories (which in this case should be considered as the derived tensor product of DG categories). It is fine to assume that the derived categories are compactly generated, e.g. when the schemes are Noetherian.
Thanks!
