Let $X$ be a noetherian scheme over base $S$ and $Y$ a closed subscheme of $X$ with arrow $j$ into $X$, $F,G$ two quasicoherent modules on $Y$. With $\boxtimes$ denote the exterior tensor product bifunctor (i.e.: pullback via the projections and tensor on the product scheme) on $Y$ resp. $X$.


Does one have a canonical iso


$(j\times j)\_\*(F \boxtimes\_Y G) \rightarrow (j\_\*F) \boxtimes\_X (j\_\*G$)?


Here the box on the left side means pullback via the projections of $Y\times\_S Y$ and on the right via the projections of $X\times\_SX$.


Or what further conditions does one need to have it?



Exterior Product of module sheaves
This is indeed true, even more general: Let $i : Y \to X$, $j : Y' \to X'$ closed immersions of $S$-schemes and $F \in \mathrm{Qcoh}(Y)$, $G \in \mathrm{Qcoh}(Y')$. Then there is a canonical isomorphism $i\_\* F \boxtimes\_{X,X'} j\_\* G \cong (i \times j)\_\* (F \boxtimes\_{Y,Y'} G)$.


Proof: The commutative diagram


$$\matrix{Y \times\_S Y' & \stackrel{i \times j}{\rightarrow} & X \times\_S X' \\\\ \mathrm{pr}\_1 \downarrow ~ & & ~ \downarrow \mathrm{pr}\_1 \\\\ Y & \stackrel{i}{\rightarrow} & X}$$


gives a canonical homomorphism $\mathrm{pr}\_1^\* i\_\* F \to (i \times j)\_\* \mathrm{pr}\_1^\* F$ (which is not an isomorphism!), similarly for $G$. Thus we get a canonical homomorphism


$$i\_\* F \boxtimes\_{X,X'} j\_\* G = \mathrm{pr}\_1^\* i\_\* F \otimes \mathrm{pr}\_2^\* j\_\* F \to (i \times j)\_\* \mathrm{pr}\_1^\* F \otimes (i \times j)\_\* \mathrm{pr}\_2^\* G$$


$$\to (i \times j)\_\* (\mathrm{pr}\_1^\* F \otimes \mathrm{pr}\_2^\* G) = (i \times j)\_\* (F \boxtimes\_{Y,Y'} G).$$


This is an isomorphism: Since we have given it globally (and its formation commutes with (co)restriction of open subsets), we may assume that everything is affine, say $S=\mathrm{Spec}(k)$, $X=\mathrm{Spec}(A)$, $X'=\mathrm{Spec}(A')$, $Y=\mathrm{Spec}(A/I)$, $Y'=\mathrm{Spec}(A'/I')$ and $F = \tilde{M}$, $G = \tilde{M'}$. Then the homomorphism looks as follows:


$$(M \otimes\_k A') \otimes\_{A \otimes\_k A'} (A \otimes\_k M') \to (M \otimes\_k A'/I') \otimes\_{A/I \otimes\_k A'/I'} (A/I \otimes\_k M')$$


which maps $(m \otimes a') \otimes (a \otimes m') \mapsto (m \otimes [a']) \otimes ([a] \otimes m')$. In order to show that it is an isomorphism, it suffices to prove that $(m \otimes i') \otimes (a \otimes m')$ vanishes in the left tensor product for all $i' \in I'$ (similarly for $I$). But this is clear, since we can rewrite this tensor as $(1 \otimes i')(m \otimes 1) \otimes (a \otimes m') = (m \otimes 1) \otimes (a \otimes i'm')$ which vanishes because $M'$ is a $A'/I'$-module.


There should be a more global proof of this (valid in a topos context etc.), but I'm too lazy to write it down.


