Box tensor product in the correspondence category

Question

I am currently reading Peter Scholze's note on six-functors formalism, where for an infinity category $C$ and a nice class of morphism $E$ in $C$, we can define the correspondence category $Corr(C,E)$ as follows:

First we define $(\triangle_{n})^{2}_{+}$ as the subset of $(\triangle^{n})^{op}\times\triangle^{n}$ spanned by $(i,j)\in \{1,\cdots,n\}^{2}$ with condition $i\geq j$, and an edge in $(\triangle_{n})^{2}_{+}$ is vertical(horizontal) if its projection in the second(first) factor is degenerate.

And the $n$ simplices of $Corr(C,E)$ are the maps from the set $(\triangle_{n})^{2}_{+}$ to $C$ with the property that

(1). All the horizontal edges correspond to morphisms in $E$. (Intuitively this means the arrows going right down are in $E$)

(2). All the squares in $C$ are Cartesian.

It can be proven that such $Corr(C,E)$ is an infinity category(see lemma 6.1.2 by Liu, Zheng), and moreover, it can be equipped with a $\textbf{symmetric monoidal structure}$ in that there is a coCartesian fibration $Corr(C,E)^{\otimes}\rightarrow N(Fin_{*})$ where the fiber over $1\in Fin_{*}$ is exactly the correspondence category $Corr(C,E)$: the concrete construction is given by $Corr(C,E)^{\otimes}=Corr((C^{op})^{\cup})^{op},E)$ where $(C^{op})^{\cup}$ is the infinity symmetric monoidal category on $C^{op}$: $(C^{op})^{\otimes}$ with coproduct symmetric monoidal structure $\otimes=\cup$ (this is product in $C$). And $E$ is also amended to be a 'nice' class of morphism in $((C^{op})^{\cup})^{op}$ which lies over the identity map in $Fin_{*}$ (so it's in the form of $f:\{X_{i\in I}\}\rightarrow \{Y_{i\in I}\}$), and for each component $i\in I$, $f_{i}\in E$.

And indeed, this symmetric monoidal category $Corr(C,E)^{\otimes}$ satisfies our intuitions in that:

(1). $Corr(C,E)^{\otimes}_{<1>}\cong Corr(C,E)$.

(2). This symmetric monoidal structure $\otimes$ is actually the $\textbf{cartesian product}$ on $C$. (recall that the objects in $Corr(C,E)$ are objects in $C$)

Point (2) is due to our construction using $^{op}$: indeed, after unravelling the definition, morphism between $(\{X_{i}\}_{i\in I},I)$ and $(\{Y_{j}\}_{j\in J},J)$ in $((C^{op})^{\cup})^{op}$ should be in the form of $\prod_{f^{-1}(j)}X_{i}\rightarrow Y_{j}$ by the universal property of the cocartesian lift. And any morphism between $(\{X_{i}\}_{i\in I},I)$ and $(\{Y_{j}\}_{j\in J},J)$ in $Corr(C,E)^{\otimes}$ should be given by $\prod_{f^{-1}(j)}X_{i}\leftarrow W_{j}\rightarrow Y_{j}$. Recall that the 'tensor product $x\otimes y$' should be given by the cocartesian lift of $<2>\rightarrow<1>\in Fin_{*}$, then by the above observation, $\{X,Y\}\rightarrow \{X\times Y\}$ is indeed 'initial' in that any composition (in the correspondence) of $X\times Y=X\times Y=X\times Y$ with $X\times Y\leftarrow W\rightarrow Z$ is again $X\times Y\leftarrow W\rightarrow Z$.

The above things can actually be $\textbf{neglected}$ just as some background of my questions.

Then there is a way of encoding 'six functors' into a $\textbf{lax symmetric monoidal functor}$ $D: Corr(C,E)^{\otimes}\rightarrow Cat_{\infty}$ (here $Cat_{\infty}$ is equipped with cartesian symmetric monoidal structure) in page 23 of Scholze's note where $\otimes:D(X)\times D(X)\rightarrow D(X)$, $f^{*}: D(Y)\rightarrow D(X)$ and $f_{!}:D(X)\rightarrow D(Y)$ along $f:X\rightarrow Y$. I think I understand this part, but later, at the top of page 24, there is a saying:

Correspondences $\prod_{i\in f^{-1}(j)}X_{i}\stackrel{f_{j}}{\longleftarrow} W_{j}\stackrel{g_{j}}{\longrightarrow} Y_{j}$ induces functor: $\prod_{I}D(X_{i})\rightarrow \prod_{J}D(Y_{j})$ whose $j$ th component is given by $(g_{j})_{!}f_{j}^{*}(\boxtimes_{i\in f^{-1}(j)})$ where the box tensor is the $\textbf{exterior tensor product}$:

$\boxtimes_{i\in f^{-1}(j)}:\prod_{i\in f^{-1}(j)}D(X_{i})\rightarrow D(\prod_{i\in f^{-1}(j)}X_{i})$ is given by tensor product of pullback.

$\textbf{Question: how to prove that the above box tensor is given by 'tensor product of pullback'?}$

I do understand in many cases, this is a very natural things, but here in the context of correspondence category, the 'tensor product' in $D(X)$ is given as $D(X\times X\leftarrow X=X)$ where the first map is induced by the diagonal morphism. What I do know about this 'box tensor' is that it's intrinsic in the definition of 'lax symmetric monoidal functor': such functor should preserve '(locally) cocartesian morphism', so by the universal property of cocartesian morphism, there is a morphism between $\otimes F(X_{i})\rightarrow F(\otimes X_{i})$, as shown above, both sides of $\otimes$ are actually cartesian product, which is our 'box tensor' above.

I believe this is a rather simple question, but I really get messed up by these giant diagrams, any suggestions or opinions are very welcome.

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$\textbf{Update:}$

So it seems that what confused me is the mysterious morphism $\boxtimes$ induced by the lax symmetric monoidal functor condition (preservation of cocartesian lift over inner morphism)): $\textbf{Can we describe it in the form of 'tensor product of pullback'?}$

This is actually a question easily leading to $\textbf{circular arguments}$ because $\otimes$ and $\boxtimes$ are actully $\textbf{determined by each other}$.

Indeed, recall the way we define the operation $\otimes$ on $D(X)$, it is the functor $D$ acting on the morphism of correspondences $(\{1,2\},(X,X))\sim (\{1\},X\times X) \stackrel{\triangle}{\longleftarrow} (\{1\},X)=(\{1\},X)$, which gives rise to $D(X)\times D(X)\stackrel{\boxtimes}{\longrightarrow}D(X\times X)\stackrel{\triangle^{*}}{\longrightarrow} D(X)$, and this composition gives us the 'tensor product operation' on $D(X)$. Indeed, $\textbf{the information about box tensor}$ $\boxtimes$ is encoded in the $\textbf{'lax symmetric monoidal structure'}$ on functor $D$, and this lax symmetric monoidal functor $D$ actually sends object $X\in C$ to the $\textbf{commutative}$ monoid object $D(X)$ in $Cat_{\infty}$ (as shown in lecture 4 of Scholze's note), whose $\textbf{'tensor operation'}$ is given by the box tensor composed with pullback of diagonal. So both the 'box tensor' and the 'tensor product' are encoded in the requirement that functor $D$ is $\textbf{lax symmetric monoidal}$.

And now we take a closer look at the above case of $D((\{1,2\},(X,X))\rightarrow (\{1\},X))$, if we have already well defined the 'tensor product' on $D(X)$ from the lax symmetric monoidal functor, then the box tensor should be exactly $\textbf{tensor product of pullback}$: indeed, $\triangle^{*}(pr_{1}^{*}(-)\otimes pr_{2}^{*}(-))=(-)\otimes (-)$ as $pr_{i}\circ \triangle=id$ (this is also true in the correspondence category). thus we have a factorization $D(X)\times D(X)\stackrel{pr_{1}(-)\otimes pr_{2}(-)}{\longrightarrow} D(X\times X)\stackrel{\triangle^{*}}{\longrightarrow}D(X)$. So intuitively speaking, if we just directly define the 'box tensor' as the 'tensor product of pullback' for the 'lax symmetric monoidal condition', we get exactly the same 'tensor product' operation as $D$ does.

The rigorous proof lies in the answer by Will Sawin, which is also a more complicated version of the above observation case. As we can see in his answer, there is a series of compositions $\prod_{i'\in f^{1}(j)}X_{i'}\leftarrow \prod_{i'\in f^{-1}(j)}((\prod_{i\in f^{-1}(j)}X_{i}))=\prod_{i'\in f^{-1}(j)}((\prod_{i\in f^{-1}(j)}X_{i}))$ (this is the product version of the first diagram in his answer), with $\prod_{i'\in f^{-1} (j)} \left(\prod_{i\in f^{-1} (j)} X_i \right) \stackrel{ \Delta}{\longleftarrow} \prod_{i\in f^{-1} (j)} X_i \stackrel{ \textrm{id}}{\longrightarrow} \prod_{i\in f^{-1} (j)} X_i$ and again with $\prod_{i\in f^{-1}(j) } X_i \stackrel{ f_j}{\longleftarrow} W_j \stackrel{\textrm{id}}{\longrightarrow} W_j$, then with $W_j \stackrel{\textrm{id}}{\longleftarrow} W_j \stackrel{g_j }{\longrightarrow} Y_j$. This whole composition of correspondences is $\textbf{identical}$ to the correspondence $\prod_{i\in f^{-1}(j)}X_{i}\stackrel{f_{j}}{\longleftarrow} W_{j}\stackrel{g_{j}}{\longrightarrow} Y_{j}$, but $\textbf{better}$, among these compositions, they $\textbf{factor through}$ the 'tensor product of pullback', which is exactly the composition of the first two morphisms of correspondences, so by this factorization, we decompose the 'mysterious box tensor' $\prod D(X_{i})\rightarrow D(\prod X_{i})$ as the 'tensor product of pullback'.

More precisely, here the expression of functor $D$ on $\prod_{i\in f^{-1}(j)}X_{i}\stackrel{f_{j}}{\longleftarrow} W_{j}\stackrel{g_{j}}{\longrightarrow} Y_{j}$ is direct from the above decomposition into smaller morphisms. Moreover, we also get the explicit expression of 'box tensor': $\prod D(X_{i})\rightarrow D(\prod X_{i})$, which is induced by the universal property of cocartesian lift. As we already get a morphism between $\prod D(X_{i})$ and $D(\prod X_{i})$, and by the above decomposition, we can see that it also satisfies the 'initial property' (because $W_{j}$ and $Y_{j}$ are actually arbitrary guys), so due to the universal property, we can say this 'box tensor' (the morphism induced by the cocartesian property) is exactly the tensor product of pullback.

Answer 1

The point is that an arbitrary correspondence can be factored as a composition of pullback, tensor, and compactly-supported pushforward correspondences.

For each $i' \in f^{-1}(j)$ we have a pullback correspondence

$$ X_{i'} \stackrel{ \pi_i}{\longleftarrow} \prod_{i\in f^{-1} (j)} X_i \stackrel{ \textrm{id}}{\longrightarrow}\prod_{i\in f^{-1} (j)} X_i $$

which gives a map in the category of correspondences from $(I, (X_{i'})_{i'\in I})$ to $(I, (\prod_{i\in f^{-1}(f(i'))} X_i )_{i'\in I)$.

and then a tensor product correspondence

$$ \prod_{i'\in f^{-1} (j)} \left(\prod_{i\in f^{-1} (j)} X_i \right) \stackrel{ \Delta}{\longleftarrow} \prod_{i\in f^{-1} (j)} X_i \stackrel{ \textrm{id}}{\longrightarrow} \prod_{i\in f^{-1} (j)} X_i $$

which gives a map in the category of correspondences from $(I, (\prod_{i\in f^{-1}(f(i'))} X_i )_{i'\in I})$ to $(J, (\prod_{i\in f^{-1}(j) } X_i )_{j\in J})$.

which composed give the $\boxtimes$ correspondence from $(I, (X_{i'})_{i'\in I})$ to $(J, (\prod_{i\in f^{-1}(j) } X_i )_{j\in J})$.

$$ \prod_{i' \in f^{-1}(j) } X_{i'} \stackrel{ i'=i}{\longleftarrow} \prod_{i \in f^{-1}(j)} X_j \stackrel{\textrm{id}}{\longrightarrow} \prod_{i \in f^{-1}(j)} X_j.$$

Note that this seems to be the identity correspondence but isn't because it's not a map from the same object to itself. For this problem, I think it's clearer to define the leftward arrows in the category of correspondences to be tuples of maps rather than (equivalently) maps to Cartesian products, but probably that causes other issues elsewhere.

If we then compose with the pullback correspondence from $(J, (\prod_{i\in f^{-1}(j) } X_i )_{j\in J})$ to $(J, (W_j)_{j\in J})$

$$ \prod_{i\in f^{-1}(j) } X_i \stackrel{ f_j}{\longleftarrow} W_j \stackrel{\textrm{id}}{\longrightarrow} W_j$$

and finally a correspondence whose rightward arrow is not $\textrm{id}$, the compactly supported pushforward correspondence

$$W_j \stackrel{\textrm{id}}{\longleftarrow} W_j \stackrel{g_j }{\longrightarrow} Y_j$$

we recover the original correspondence, showing the description of $D$ of the original correspondence as a composition of compactly supported pushforward, pullback, and boxtimes which is itself a composition of tensor product and pullback is correct.

If necessary I can explain how to calculate the compositions of these arrows. One can calculate the compositions two at a time, which requires finding a single Cartesian square and then composing, and then in most cases one of the morphisms in the square is an identity morphism and so one just has to compose.