4.6.4 A Six-Functor Formalism for Sets

The assignment $X\mapsto \mathcal{P}(X)$ together with the functors $f_{!}$, $f^{-1}$, and $f_{*}$ of Item 1 of Proposition 4.6.1.1.5, Item 1 of Proposition 4.6.2.1.3, and Item 1 of Proposition 4.6.3.1.7, and the functors

\begin{align*} -_{1}\cap -_{2} & \colon \mathcal{P}(X)\times \mathcal{P}(X) \to \mathcal{P}(X),\\ [-_{1},-_{2}]_{X} & \colon \mathcal{P}(X)^{\mathsf{op}}\times \mathcal{P}(X) \to \mathcal{P}(X) \end{align*}

of Item 1 of Proposition 4.3.9.1.2 and Item 1 of Proposition 4.4.7.1.3 satisfy several properties reminiscent of a six functor formalism in the sense of Unresolved reference.

We collect these properties in Proposition 4.6.4.1.2 below.1

Let $X$ be a set.

  1. 1.

    The Beck–Chevalley Condition. Let

    be a pullback diagram in $\mathsf{Sets}$. We have

  2. 2.

    The Projection Formula I. The diagram

    commutes, i.e. we have

    \[ f_{!}(U\cap f^{-1}(V))=f_{!}(U)\cap V \]

    for each $U\in \mathcal{P}(X)$ and each $V\in \mathcal{P}(Y)$.

  3. 3.

    The Projection Formula II. We have a natural transformation

    with components

    \[ f_{*}(U)\cap V\subset f_{*}(U\cap f^{-1}(V)) \]

    indexed by $U\in \mathcal{P}(X)$ and $V\in \mathcal{P}(Y)$.

  4. 4.

    Strong Closed Monoidality. The diagram

    commutes, i.e. we have an equality of sets

    \[ f^{-1}([U,V]_{Y})=[f^{-1}(U),f^{-1}(V)]_{X}, \]

    natural in $U,V\in \mathcal{P}(X)$.

  5. 5.

    The External Tensor Product. We have an external tensor product

    \[ -_{1}\boxtimes _{X\times Y}-_{2}\colon \mathcal{P}(X)\times \mathcal{P}(Y)\to \mathcal{P}(X\times Y) \]

    given by

    \begin{align*} U\boxtimes _{X\times Y}V & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{\mathrm{pr}}}^{-1}_{1}(U)\cap \operatorname {\mathrm{\mathrm{pr}}}^{-1}_{2}(V)\\ & = \left\{ (u,v)\in X\times Y\ \middle |\ \text{$u\in U$ and $v\in V$}\right\} . \end{align*}

    This is the same map as the one in Item 5 of Proposition 4.4.1.1.4. Moreover, the following conditions are satisfied:

    1. (a)

      Interaction With Direct Images. Let $f\colon X\to X'$ and $g\colon Y\to Y'$ be functions. The diagram

      commutes, i.e. we have

      \[ [f_{!}\times g_{!}](U\boxtimes _{X\times Y}V)=f_{!}(U)\boxtimes _{X'\times Y'}g_{!}(V) \]

      for each $(U,V)\in \mathcal{P}(X)\times \mathcal{P}(Y)$.

    2. (b)

      Interaction With Inverse Images. Let $f\colon X\to X'$ and $g\colon Y\to Y'$ be functions. The diagram

      commutes, i.e. we have

      \[ [f^{-1}\times g^{-1}](U\boxtimes _{X'\times Y'}V)=f^{-1}(U)\boxtimes _{X\times Y}g^{-1}(V) \]

      for each $(U,V)\in \mathcal{P}(X')\times \mathcal{P}(Y')$.

    3. (c)

      Interaction With Codirect Images. Let $f\colon X\to X'$ and $g\colon Y\to Y'$ be functions. The diagram

      commutes, i.e. we have

      \[ [f_{*}\times g_{*}](U\boxtimes _{X\times Y}V)=f_{*}(U)\boxtimes _{X'\times Y'}g_{*}(V) \]

      for each $(U,V)\in \mathcal{P}(X)\times \mathcal{P}(Y)$.

    4. (d)

      Interaction With Diagonals. The diagram

      i.e. we have

      \[ U\cap V=\Delta ^{-1}_{X}(U\boxtimes _{X\times X}V) \]

      for each $U,V\in \mathcal{P}(X)$.

  6. 6.

    The Dualisation Functor. We have a functor

    \[ D_{X}\colon \mathcal{P}(X)^{\mathsf{op}}\to \mathcal{P}(X) \]

    given by

    \begin{align*} D_{X}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[U,\text{Ø}]_{X}\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}U^{\textsf{c}} \end{align*}

    for each $U\in \mathcal{P}(X)$, as in Item 5 of Proposition 4.4.7.1.3, satisfying the following conditions:

    1. (a)

      Duality. We have

    2. (b)

      Interaction With Internal Homs. The diagram

      commutes, i.e. we have

      \[ \underbrace{D_{X}(U\cap D_{X}(V))}_{\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[U\cap [V,\text{Ø}]_{X},\text{Ø}]_{X}}=[U,V]_{X} \]

      for each $U,V\in \mathcal{P}(X)$.

    3. (c)

      Interaction With Direct Images. The diagram

      commutes, i.e. we have

      \[ f_{!}(D_{X}(U))=D_{Y}(f_{*}(U)) \]

      for each $U\in \mathcal{P}(X)$.

    4. (d)

      Interaction With Inverse Images. The diagram

      commutes, i.e. we have

      \[ f^{-1}(D_{Y}(U))=D_{X}(f^{-1}(U)) \]

      for each $U\in \mathcal{P}(X)$.

    5. (e)

      Interaction With Codirect Images. The diagram

      commutes, i.e. we have

      \[ f_{*}(D_{X}(U))=D_{Y}(f_{!}(U)) \]

      for each $U\in \mathcal{P}(X)$.

Item 1: The Beck–Chevalley Condition
We have

\begin{align*} [g^{-1}\circ f_{!}](U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}g^{-1}(f_{!}(U))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ y\in Y\ \middle |\ g(y)\in f_{!}(U)\right\} \\ & = \left\{ y\in Y\ \middle |\ \begin{aligned} & \text{there exists some $x\in U$}\\ & \text{such that $f(x)=g(y)$}\end{aligned}\right\} \\ & = \left\{ y\in Y\ \middle |\ \begin{aligned} & \text{there exists some}\\ & \text{$(x,y)\in \left\{ (x,y)\in X\times _{Z}Y\ \middle |\ x\in U\right\} $}\end{aligned}\right\} \\ & = \left\{ y\in Y\ \middle |\ \begin{aligned} & \text{there exists some}\\ & \text{$(x,y)\in \left\{ (x,y)\in X\times _{Z}Y\ \middle |\ x\in U\right\} $}\\ & \text{such that $y=y$}\end{aligned}\right\} \\ & = \left\{ y\in Y\ \middle |\ \begin{aligned} & \text{there exists some}\\ & \text{$(x,y)\in \left\{ (x,y)\in X\times _{Z}Y\ \middle |\ x\in U\right\} $}\\ & \text{such that $\operatorname {\mathrm{\mathrm{pr}}}_{2}(x,y)=y$}\end{aligned}\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(\operatorname {\mathrm{\mathrm{pr}}}_{2})_{!}(\left\{ (x,y)\in X\times _{Z}Y\ \middle |\ x\in U\right\} )\\ & = (\operatorname {\mathrm{\mathrm{pr}}}_{2})_{!}(\left\{ (x,y)\in X\times _{Z}Y\ \middle |\ \operatorname {\mathrm{\mathrm{pr}}}_{1}(x,y)\in U\right\} )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(\operatorname {\mathrm{\mathrm{pr}}}_{2})_{!}(\operatorname {\mathrm{\mathrm{pr}}}^{-1}_{1}(U))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[(\operatorname {\mathrm{\mathrm{pr}}}_{2})_{!}\circ \operatorname {\mathrm{\mathrm{pr}}}^{-1}_{1}](U) \end{align*}

for each $U\in \mathcal{P}(X)$. Therefore, we have

\[ g^{-1}\circ f_{!}=(\operatorname {\mathrm{\mathrm{pr}}}_{2})_{!}\circ \operatorname {\mathrm{\mathrm{pr}}}^{-1}_{1}. \]

For the second equality, we have

\begin{align*} [f^{-1}\circ g_{!}](U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f^{-1}(g_{!}(U))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ f(x)\in g_{!}(V)\right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some $y\in V$}\\ & \text{such that $f(x)=g(y)$}\end{aligned}\right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some}\\ & \text{$(x,y)\in \left\{ (x,y)\in X\times _{Z}Y\ \middle |\ y\in V\right\} $}\end{aligned}\right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some}\\ & \text{$(x,y)\in \left\{ (x,y)\in X\times _{Z}Y\ \middle |\ y\in V\right\} $}\\ & \text{such that $x=x$}\end{aligned}\right\} \\ & = \left\{ x\in X\ \middle |\ \begin{aligned} & \text{there exists some}\\ & \text{$(x,y)\in \left\{ (x,y)\in X\times _{Z}Y\ \middle |\ y\in V\right\} $}\\ & \text{such that $\operatorname {\mathrm{\mathrm{pr}}}_{1}(x,y)=x$}\end{aligned}\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(\operatorname {\mathrm{\mathrm{pr}}}_{1})_{!}(\left\{ (x,y)\in X\times _{Z}Y\ \middle |\ y\in V\right\} )\\ & = (\operatorname {\mathrm{\mathrm{pr}}}_{1})_{!}(\left\{ (x,y)\in X\times _{Z}Y\ \middle |\ \operatorname {\mathrm{\mathrm{pr}}}_{2}(x,y)\in V\right\} )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(\operatorname {\mathrm{\mathrm{pr}}}_{1})_{!}(\operatorname {\mathrm{\mathrm{pr}}}^{-1}_{2}(V))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[(\operatorname {\mathrm{\mathrm{pr}}}_{1})_{!}\circ \operatorname {\mathrm{\mathrm{pr}}}^{-1}_{2}](V) \end{align*}

for each $V\in \mathcal{P}(Y)$. Therefore, we have

\[ f^{-1}\circ g_{!}=(\operatorname {\mathrm{\mathrm{pr}}}_{1})_{!}\circ \operatorname {\mathrm{\mathrm{pr}}}^{-1}_{2}. \]

This finishes the proof.

Item 2: The Projection Formula I
We claim that

\[ f_{!}(U)\cap V\subset f_{!}(U\cap f^{-1}(V)). \]

Indeed, we have

\begin{align*} f_{!}(U)\cap V & \subset f_{!}(U)\cap f_{!}(f^{-1}(V))\\ & = f_{!}(U\cap f^{-1}(V)), \end{align*}

where we have used:

  1. 1.

    Item 2 of Proposition 4.6.1.1.5 for the inclusion.

  2. 2.

    Item 6 of Proposition 4.6.1.1.5 for the equality.

Conversely, we claim that

\[ f_{!}(U\cap f^{-1}(V))\subset f_{!}(U)\cap V. \]

Indeed:

  1. 1.

    Let $y\in f_{!}(U\cap f^{-1}(V))$.

  2. 2.

    Since $y\in f_{!}(U\cap f^{-1}(V))$, there exists some $x\in U\cap f^{-1}(V)$ such that $f(x)=y$.

  3. 3.

    Since $x\in U\cap f^{-1}(V)$, we have $x\in U$, and thus $f(x)\in f_{!}(U)$.

  4. 4.

    Since $x\in U\cap f^{-1}(V)$, we have $x\in f^{-1}(V)$, and thus $f(x)\in V$.

  5. 5.

    Since $f(x)\in f_{!}(U)$ and $f(x)\in V$, we have $f(x)\in f_{!}(U)\cap V$.

  6. 6.

    But $y=f(x)$, so $y\in f_{!}(U)\cap V$.

  7. 7.

    Thus $f_{!}(U\cap f^{-1}(V))\subset f_{!}(U)\cap V$.

This finishes the proof.

Item 3: The Projection Formula II
We have

\begin{align*} f_{*}(U)\cap V & \subset f_{*}(U)\cap f_{*}(f^{-1}(V))\\ & = f_{*}(U\cap f^{-1}(V)), \end{align*}

where we have used:

  1. 1.

    Item 2 of Proposition 4.6.3.1.7 for the inclusion.

  2. 2.

    Item 6 of Proposition 4.6.3.1.7 for the equality.

Since $\mathcal{P}(Y)$ is posetal, naturality is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2).

Item 4: Strong Closed Monoidality
This is a repetition of Item 19 of Proposition 4.4.7.1.3 and is proved there.

Item 5: The External Tensor Product
We have

\begin{align*} U\boxtimes _{X\times Y}V & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{\mathrm{pr}}}^{-1}_{1}(U)\cap \operatorname {\mathrm{\mathrm{pr}}}^{-1}_{2}(V)\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ (x,y)\in X\times Y\ \middle |\ \operatorname {\mathrm{\mathrm{pr}}}_{1}(x,y)\in U\right\} \\ & \phantom{={}} \mkern 4mu\cup \left\{ (x,y)\in X\times Y\ \middle |\ \operatorname {\mathrm{\mathrm{pr}}}_{2}(x,y)\in V\right\} \\ & = \left\{ (x,y)\in X\times Y\ \middle |\ x\in U\right\} \\ & \phantom{={}} \mkern 4mu\cup \left\{ (x,y)\in X\times Y\ \middle |\ y\in V\right\} \\ & = \left\{ (x,y)\in X\times Y\ \middle |\ \text{$x\in U$ and $y\in V$}\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}U\times V. \end{align*}

Next, we claim that Item 5a, Item 5b, Item 5c, and Item 5d are indeed true:

  1. 1.

    Proof of Item 5a: This is a repetition of Item 16 of Proposition 4.6.1.1.5 and is proved there.

  2. 2.

    Proof of Item 5b: This is a repetition of Item 16 of Proposition 4.6.2.1.3 and is proved there.

  3. 3.

    Proof of Item 5c: This is a repetition of Item 15 of Proposition 4.6.3.1.7 and is proved there.

  4. 4.

    Proof of Item 5d: We have

    \begin{align*} \Delta ^{-1}_{X}(U\boxtimes _{X\times X}V) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ x\in X\ \middle |\ (x,x)\in U\boxtimes _{X\times X}V\right\} \\ & = \left\{ x\in X\ \middle |\ (x,x)\in \left\{ (u,v)\in X\times X\ \middle |\ \text{$u\in U$ and $v\in V$}\right\} \right\} \\ & = U\cap V.\end{align*}

This finishes the proof.

Item 6: The Dualisation Functor
This is a repetition of Item 5 and Item 6 of Proposition 4.4.7.1.3 and is proved there.


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