Omitted.
This follows from
,
of
.
Item 3: Interaction With Unions of Families of Subsets
We have
\begin{align*} \bigcup _{U\in R_{*}(\mathcal{V})}U & = \bigcup _{U\in \left\{ R_{*}(V)\in \mathcal{P}(X)\ \middle |\ V\in \mathcal{V}\right\} }U\\ & = \bigcup _{V\in \mathcal{V}}R_{*}(V).\end{align*}
This finishes the proof.
Item 4: Interaction With Intersections of Families of Subsets
We have
\begin{align*} \bigcap _{U\in R_{*}(\mathcal{V})}U & = \bigcap _{U\in \left\{ R_{*}(V)\in \mathcal{P}(X)\ \middle |\ V\in \mathcal{V}\right\} }U\\ & = \bigcap _{V\in \mathcal{V}}R_{*}(V).\end{align*}
This finishes the proof.
Item 5: Interaction With Binary Unions
We have
\begin{align*} R_{*}(U)\cup R_{*}(V) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ b\in Y\ \middle |\ R^{-1}(b)\subset U\right\} \cup \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset V\right\} \\ & = \left\{ b\in Y\ \middle |\ \text{$R^{-1}(b)\subset U$ or $R^{-1}(b)\subset V$}\right\} \\ & \subset \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset U\cup V\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R_{-1}(U\cup V). \end{align*}
Since $\mathcal{P}(X)$ is posetal, naturality is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2). This finishes the proof.
Item 6: Interaction With Binary Intersections
We have
\begin{align*} R_{*}(U)\cap R_{*}(V) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ b\in X\ \middle |\ R^{-1}(b)\subset U\right\} \cap \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset V\right\} \\ & = \left\{ b\in Y\ \middle |\ \text{$R^{-1}(b)\subset U$ and $R^{-1}(b)\subset V$}\right\} \\ & = \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset U\cap V\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R_{*}(U\cap V). \end{align*}
This finishes the proof.
Item 7: Interaction With Differences
We have
\begin{align*} R_{*}(U\setminus V) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ b\in Y\ \middle |\ R^{-1}(b)\subset U\setminus V\right\} \\ & \subset \left\{ b\in Y\ \middle |\ \text{$R^{-1}(b)\subset U$ and $R^{-1}(b)\centernot {\subset }V$}\right\} \\ & = \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset U\right\} \setminus \left\{ b\in Y\ \middle |\ R^{-1}(b)\subset V\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R_{*}(U)\setminus R_{*}(V)\\ \end{align*}
Since $\mathcal{P}(X)$ is posetal, naturality is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2). This finishes the proof.
Item 8: Interaction With Complements
Applying Item 17 to $X\setminus U$, we have
\begin{align*} R_{*}(U^{\textsf{c}}) & = R_{*}(X\setminus U)\\ & = Y\setminus R_{!}(X\setminus (X\setminus U))\\ & = Y\setminus R_{!}(U)\\ & = R_{!}(U)^{\textsf{c}}. \end{align*}
This finishes the proof.
Item 9: Interaction With Symmetric Differences
Omitted.
Item 10: Interaction With Internal Homs of Powersets
We have
\begin{align*} [R_{!}(U),R_{*}(V)]_{Y} & = R_{!}(U)^{\textsf{c}}\cup R_{*}(V)\\ & = R_{*}(U^{\textsf{c}})\cup R_{*}(V)\\ & \subset R_{*}(U^{\textsf{c}}\cup V)\\ & = R_{*}([U,V]_{X}).\end{align*}
Since $\mathcal{P}(X)$ is posetal, naturality is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2). This finishes the proof.
Item 11: Lax Preservation of Colimits
Omitted.
This follows from Item 2 and
,
of
.
Item 13: Symmetric Lax Monoidality With Respect to Unions
This follows from Item 11.
Item 14: Symmetric Strict Monoidality With Respect to Intersections
This follows from Item 12.
Item 15: Interaction With Coproducts
Omitted.
Omitted.
This follows from Item 17 of Proposition 8.7.1.1.4. Alternatively, we may prove it directly as follows, with the proof proceeding in the same way as in the case of functions (Chapter 4: Constructions With Sets, Item 16 of Proposition 4.6.3.1.7).
We claim that $R_{*}(U)=Y\setminus R_{!}(X\setminus U)$:
-
•
Proof of $R_{*}(U)\subset Y\setminus R_{!}(X\setminus U)$: We proceed in a few steps:
-
•
Let $b\in R_{*}(U)$. We need to show that $b\not\in R_{!}(X\setminus U)$, i.e. that there is no $a\in X\setminus U$ such that $b\in R(a)$.
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•
This is indeed the case, as otherwise we would have $a\in R^{-1}(b)$ and $a\not\in U$, contradicting $R^{-1}(b)\subset U$ (which holds since $b\in R_{*}(U)$).
-
•
Thus $b\in Y\setminus R_{!}(X\setminus U)$.
-
•
Proof of $Y\setminus R_{!}(X\setminus U)\subset R_{*}(U)$: We proceed in a few steps:
-
•
Let $b\in Y\setminus R_{!}(X\setminus U)$. We need to show that $b\in R_{*}(U)$, i.e. that $R^{-1}(b)\subset U$.
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•
Since $b\not\in R_{!}(X\setminus U)$, there exists no $a\in X\setminus U$ such that $b\in R(a)$, and hence $R^{-1}(b)\subset U$.
-
•
Thus $b\in R_{*}(U)$.
This finishes the proof.
Item 18: Interaction With Injections
If $R$ is injective, then the condition $R^{-1}(y)\subset U$ is equivalent to $R^{-1}(y)\cap U\neq \text{Ø}$ when $y\in \mathrm{Im}(R)$. Thus $R^{\mathrm{im}}_{*}(U)=R_{!}(U)$.
Item 19: Interaction With Surjections
Indeed, we have
\begin{align*} R^{\mathrm{cp}}_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}Y\setminus \mathrm{Im}(R)\\ & = Y\setminus Y\\ & = \text{Ø}. \end{align*}
Since $R^{-1}(y)\subset U$ is a strictly stronger condition than $R^{-1}(y)\cap U\neq \text{Ø}$ when $R^{-1}(y)$ is nonempty, we have $R^{\mathrm{im}}_{*}(U)\subset R_{!}(U)$. Thus $R_{*}(U)\subset R_{!}(U)$.
Item 20: Interaction With Bijections I
This follows from Item 18 and Item 19.
Item 21: Interaction With Bijections II
First, recall from
that we have
\[ R_{*}(U)=Y\setminus R_{!}(X\setminus U) \]
for all $U\in \mathcal{P}(A)$. We now claim that $R$ is bijective:
-
•
Injectivity: We proceed in a few steps:
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•
Choosing $U=\left\{ a\right\} $, we obtain $R(a)=Y\setminus R(X\setminus \left\{ a\right\} )$.
-
•
This implies $R(a)\cap R(X\setminus \left\{ a\right\} )=\text{Ø}$.
-
•
Thus there can be no $a'\in X$ other than $a$ with $R(a')\cap R(a')\neq \text{Ø}$.
-
•
Surjectivity: Choosing $U=X$ gives $R_{*}(X)=Y$. But since $R_{!}=R_{*}$, we have $R_{!}(X)=Y$, so $R$ must be surjective.
This finishes the proof.
There is nothing to prove.
This follows from Item 1 of Proposition 8.7.4.1.4.
Item 3: Interaction With Identities
Indeed, we have
\begin{align*} (\chi _{X})_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ a\in X\ \middle |\ \chi ^{-1}_{X}(a)\subset U\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ a\in X\ \middle |\ \left\{ a\right\} \subset U\right\} \\ & = U \end{align*}
for each $U\in \mathcal{P}(X)$. Thus $(\chi _{X})_{*}=\operatorname {\mathrm{id}}_{\mathcal{P}(X)}$.
Item 4: Interaction With Composition
Indeed, we have
\begin{align*} (S\mathbin {\diamond }R)_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ c\in C\ \middle |\ [S\mathbin {\diamond }R]^{-1}(c)\subset U\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ c\in C\ \middle |\ S^{-1}(R^{-1}(c))\subset U\right\} \\ & = \left\{ c\in C\ \middle |\ R^{-1}(c)\subset S_{*}(U)\right\} \\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}R_{*}(S_{*}(U))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[R_{*}\circ S_{*}](U) \end{align*}
for each $U\in \mathcal{P}(C)$, where we used Item 2 of Proposition 8.7.4.1.4, which implies that the conditions
-
•
We have $S^{-1}(R^{-1}(c))\subset U$.
-
•
We have $R^{-1}(c)\subset S_{*}(U)$.
are equivalent. Thus $(S\mathbin {\diamond }R)_{*}=S_{*}\circ R_{*}$.
Item 5: Interaction With Converses
Omitted.