4.6.3 Codirect Images
Let $f\colon X\to Y$ be a function.
The codirect image function associated to $f$ is the function
\[ f_{*}\colon \mathcal{P}(X)\to \mathcal{P}(Y) \]
defined by
\begin{align*} f_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ y\in Y\ \middle |\ \begin{aligned} & \text{for each $x\in X$, if we have}\\ & \text{$f(x)=y$, then $x\in U$}\end{aligned} \right\} \\ & = \left\{ y\in Y\ \middle |\ \text{we have $f^{-1}(y)\subset U$}\right\} \end{align*}
for each $U\in \mathcal{P}(X)$.
Sometimes one finds the notation
\[ \forall _{f}\colon \mathcal{P}(X)\to \mathcal{P}(Y) \]
for $f_{!}$. This notation comes from the fact that the following statements are equivalent, where $y\in Y$ and $U\in \mathcal{P}(X)$:
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We have $y\in \forall _{f}(U)$.
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For each $x\in X$, if $y=f(x)$, then $x\in U$.
We will not make use of this notation elsewhere in Clowder.
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See Warning 4.6.1.1.3.
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Let $U$ be a subset of $X$.
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The image part of the codirect image $f_{*}(U)$ of $U$ is the set $f^{\mathrm{im}}_{*}(U)$ defined by
\begin{align*} f^{\mathrm{im}}_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f_{*}(U)\cap \mathrm{Im}(f)\\ & = \left\{ y\in Y\ \middle |\ \begin{aligned} & \text{we have $f^{-1}(y)\subset U$}\\ & \text{and $f^{-1}(y)\neq \text{Ø}$.} \end{aligned}\right\} .\end{align*}
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The complement part of the codirect image $f_{*}(U)$ of $U$ is the set $f^{\mathrm{cp}}_{*}(U)$ defined by
\begin{align*} f^{\mathrm{cp}}_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f_{*}(U)\cap (Y\setminus \mathrm{Im}(f))\\ & = Y\setminus \mathrm{Im}(f)\\ & = \left\{ y\in Y\ \middle |\ \begin{aligned} & \text{we have $f^{-1}(y)\subset U$}\\ & \text{and $f^{-1}(y)=\text{Ø}$.} \end{aligned}\right\} \\ & = \left\{ y\in Y\ \middle |\ f^{-1}(y)=\text{Ø}\right\} .\end{align*}
Here are some examples of codirect images.
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Multiplication by Two. Consider the function $f\colon \mathbb {N}\to \mathbb {N}$ given by
\[ f(n) \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}2n \]
for each $n\in \mathbb {N}$. Since $f$ is injective, we have
\begin{align*} f^{\mathrm{im}}_{*}(U) & = f_{!}(U)\\ f^{\mathrm{cp}}_{*}(U) & = \left\{ \text{odd natural numbers}\right\} \end{align*}
for any $U\subset \mathbb {N}$. In particular, we have
\[ f_{*}(\left\{ \text{even natural numbers}\right\} )=\mathbb {N}. \]
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Parabolas. Consider the function $f\colon \mathbb {R}\to \mathbb {R}$ given by
\[ f(x)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x^{2} \]
for each $x\in \mathbb {R}$. We have
\[ f^{\mathrm{cp}}_{*}(U)=\mathbb {R}_{<0} \]
for any $U\subset \mathbb {R}$. Moreover, since $f^{-1}(x)=\left\{ -\sqrt{x},\sqrt{x}\right\} $, we have e.g.:
\begin{gather*} \begin{aligned} f^{\mathrm{im}}_{*}([0,1]) & = \left\{ 0\right\} ,\\ f^{\mathrm{im}}_{*}([-1,1]) & = [0,1],\\ f^{\mathrm{im}}_{*}([1,2]) & = \text{Ø}, \end{aligned}\\ f^{\mathrm{im}}_{*}([-2,-1]\cup [1,2]) = [1,4]. \end{gather*}
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Circles. Consider the function $f\colon \mathbb {R}^{2}\to \mathbb {R}$ given by
\[ f(x,y)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x^{2}+y^{2} \]
for each $(x,y)\in \mathbb {R}^{2}$. We have
\[ f^{\mathrm{cp}}_{*}(U)=\mathbb {R}_{<0} \]
for any $U\subset \mathbb {R}^{2}$, and since
\[ f^{-1}(r)= \begin{cases} \text{a circle of radius $r$ about the origin} & \text{if $r>0$,}\\ \left\{ (0,0)\right\} & \text{if $r=0$,}\\ \text{Ø}& \text{if $r<0$,} \end{cases} \]
we have e.g.:
\begin{gather*} f^{\mathrm{im}}_{*}([-1,1]\times [-1,1]) = [0,1],\\ f^{\mathrm{im}}_{*}(([-1,1]\times [-1,1])\setminus [-1,1]\times \left\{ 0\right\} ) = \text{Ø}. \end{gather*}
Omitted.
This follows from Remark 4.6.1.1.4, Remark 4.6.2.1.2, Remark 4.6.3.1.4, and
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Item 3: Interaction With Unions of Families of Subsets
We have
\begin{align*} \bigcup _{V\in f_{*}(\mathcal{U})}V & = \bigcup _{V\in \left\{ f_{*}(U)\in \mathcal{P}(X)\ \middle |\ U\in \mathcal{U}\right\} }V\\ & = \bigcup _{U\in \mathcal{U}}f_{*}(U).\end{align*}
This finishes the proof.
Item 4: Interaction With Intersections of Families of Subsets
We have
\begin{align*} \bigcap _{V\in f_{*}(\mathcal{U})}V & = \bigcap _{V\in \left\{ f_{*}(U)\in \mathcal{P}(X)\ \middle |\ U\in \mathcal{U}\right\} }V\\ & = \bigcap _{U\in \mathcal{U}}f_{*}(U).\end{align*}
This finishes the proof.
Item 5: Interaction With Binary Unions
We have
\begin{align*} f_{*}(U)\cup f_{*}(V) & = f_{!}(U^{\textsf{c}})^{\textsf{c}}\cup f_{!}(V^{\textsf{c}})^{\textsf{c}}\\ & = (f_{!}(U^{\textsf{c}})\cap f_{!}(V^{\textsf{c}}))^{\textsf{c}}\\ & \subset (f_{!}(U^{\textsf{c}}\cap V^{\textsf{c}}))^{\textsf{c}}\\ & = f_{!}((U\cup V)^{\textsf{c}})^{\textsf{c}}\\ & = f_{*}(U\cup V), \end{align*}
where:
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We have used Item 16 for the first equality.
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We have used Item 2 of Proposition 4.3.11.1.2 for the second equality.
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We have used Item 6 of Proposition 4.6.1.1.5 for the third equality.
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We have used Item 2 of Proposition 4.3.11.1.2 for the fourth equality.
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We have used Item 16 for the last equality.
This finishes the proof.
Item 6: Interaction With Binary Intersections
This follows from Item 11.
Item 7: Interaction With Complements
Omitted.
Item 8: Interaction With Symmetric Differences
Omitted.
Item 9: Interaction With Internal Homs of Powersets
We have
\begin{align*} \big [f_{!}(U),f^{!}(V)\big ]_{X} & = f_{!}(U)^{\textsf{c}}\cup f_{*}(V)\\ & = f_{*}(U^{\textsf{c}})\cup f_{*}(V)\\ & \subset f_{*}(U^{\textsf{c}}\cup V)\\ & = f_{*}([U,V]_{X}), \end{align*}
where we have used:
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Item 7 of Proposition 4.6.3.1.7 for the second equality.
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Item 5 of Proposition 4.6.3.1.7 for the inclusion.
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 10: Lax Preservation of Colimits
Omitted.
This follows from Item 2 and
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Item 12: Symmetric Lax Monoidality With Respect to Unions
This follows from Item 10.
Item 13: Symmetric Strict Monoidality With Respect to Intersections
This follows from Item 11.
Item 14: Interaction With Coproducts
Omitted.
Omitted.
We claim that $f_{*}(U)=Y\setminus f_{!}(X\setminus U)$.
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The First Implication. We claim that
\[ f_{*}(U)\subset Y\setminus f_{!}(X\setminus U). \]
Let $y\in f_{*}(U)$. We need to show that $y\not\in f_{!}(X\setminus U)$, i.e. that there is no $x\in X\setminus U$ such that $f(x)=y$.
This is indeed the case, as otherwise we would have $x\in f^{-1}(y)$ and $x\not\in U$, contradicting $f^{-1}(y)\subset U$ (which holds since $y\in f_{*}(U)$).
Thus $y\in Y\setminus f_{!}(X\setminus U)$.
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The Second Implication. We claim that
\[ Y\setminus f_{!}(X\setminus U)\subset f_{*}(U). \]
Let $y\in Y\setminus f_{!}(X\setminus U)$. We need to show that $y\in f_{*}(U)$, i.e. that $f^{-1}(y)\subset U$.
Since $y\not\in f_{!}(X\setminus U)$, there exists no $x\in X\setminus U$ such that $y=f(x)$, and hence $f^{-1}(y)\subset U$.
Thus $y\in f_{*}(U)$.
This finishes the proof of Item 16.
Item 17: Interaction With Injections
If $f$ is injective, then the condition $f^{-1}(y)\subset U$ is equivalent to $f^{-1}(y)\cap U\neq \text{Ø}$ when $y\in \mathrm{Im}(f)$. Thus $f^{\mathrm{im}}_{*}(U)=f_{!}(U)$.
Item 18: Interaction With Surjections
Indeed, we have
\begin{align*} f^{\mathrm{cp}}_{*}(U) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}Y\setminus \mathrm{Im}(f)\\ & = Y\setminus Y\\ & = \text{Ø}. \end{align*}
Since $f^{-1}(y)\subset U$ is a strictly stronger condition than $f^{-1}(y)\cap U\neq \text{Ø}$ when $f^{-1}(y)$ is nonempty, we have $f^{\mathrm{im}}_{*}(U)\subset f_{!}(U)$. Thus $f_{*}(U)\subset f_{!}(U)$.
Item 19: Interaction With Bijections I
This follows from Item 17 and Item 18.
Item 20: Interaction With Bijections II
First, recall from Item 16 that we have
\[ f_{*}(U)=Y\setminus f_{!}(X\setminus U) \]
for all $U\in \mathcal{P}(A)$. We now claim that $f$ is bijective:
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Injectivity: We proceed in a few steps:
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Choosing $U=\left\{ a\right\} $, we obtain $\left\{ f(a)\right\} =Y\setminus f(X\setminus \left\{ a\right\} )$.
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This implies $\left\{ f(a)\right\} \cap f(X\setminus \left\{ a\right\} )=\text{Ø}$.
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Thus there can be no $a'\in X$ other than $a$ with $f(a')=f(a)$.
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Surjectivity: Choosing $U=X$ gives $f_{*}(X)=Y$. But since $f_{!}=f_{*}$, we have $f_{!}(X)=Y$, so $f$ must be surjective.
This finishes the proof.