4.6.1 Direct Images
Let $f\colon X\to Y$ be a function.
The direct 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{there exists some $x\in U$}\\ & \text{such that $y=f(x)$} \end{aligned} \right\} \\ & = \left\{ y\in Y\ \middle |\ f^{-1}(y)\cap U\neq \text{Ø}\right\} \\ & = \left\{ f(x)\in Y\ \middle |\ x\in U\right\} \end{align*}
for each $U\in \mathcal{P}(X)$.
Sometimes one finds the notation
\[ \exists _{f}\colon \mathcal{P}(X)\to \mathcal{P}(Y) \]
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 $y\in \exists _{f}(U)$.
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There exists some $x\in U$ such that $f(x)=y$.
We will not make use of this notation elsewhere in Clowder.
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Notation for direct images between powersets is tricky:
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1.
Direct images for powersets and presheaves are both adjoint to their corresponding inverse image functors. However, the direct image functor for powersets is a left adjoint, while the direct image functor for presheaves is a right adjoint:
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(a)
Powersets. Given a function $f\colon X\to Y$, we have an inverse image functor
\[ f^{-1}\colon \mathcal{P}(Y)\to \mathcal{P}(X). \]
The left adjoint of this functor is the usual direct image, defined above in Definition 4.6.1.1.1.
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(b)
Presheaves. Given a morphism of topological spaces $f\colon X\to Y$, we have an inverse image functor
\[ f^{-1}\colon \mathsf{PSh}(Y)\to \mathsf{PSh}(X). \]
The right adjoint of this functor is the direct image functor of presheaves, defined in
.
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2.
The presheaf direct image functor is denoted $f_{*}$, but the direct image functor for powersets is denoted $f_{!}$ (as it’s a left adjoint).
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3.
Adding to the confusion, it’s somewhat common for $f_{!}\colon \mathcal{P}(X)\to \mathcal{P}(Y)$ to be denoted $f_{*}$.
We chose to write $f_{!}$ for the direct image to keep the notation aligned with the following similar adjoint situations:
Situation
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Adjoint String
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Functoriality
of Powersets
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$(f_{!}\dashv f^{-1}\dashv f_{*})\colon \mathcal{P}(X)\overset {\rightleftarrows }{\to }\mathcal{P}(Y)$
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Functoriality of
Presheaf Categories
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$(f_{!}\dashv f^{-1}\dashv f_{*})\colon \mathsf{PSh}(X)\overset {\rightleftarrows }{\to }\mathsf{PSh}(Y)$
|
Base Change
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$(f_{!}\dashv f^{*}\dashv f_{*})\colon \mathcal{C}_{/X}\overset {\rightleftarrows }{\to }\mathcal{C}_{/Y}$
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Kan Extensions
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$(F_{!}\dashv F^{*}\dashv F_{*})\colon \mathsf{Fun}(\mathcal{C},\mathcal{E})\overset {\rightleftarrows }{\to }\mathsf{Fun}(\mathcal{D},\mathcal{E})$
|
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Omitted.
This follows from Remark 4.6.1.1.4, Remark 4.6.2.1.2, Remark 4.6.3.1.4, and
,
of
.
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
See [Proof Wiki Contributors, Image of Union Under Mapping — Proof Wiki].
Item 6: Interaction With Binary Intersections
See [Proof Wiki Contributors, Image of Intersection Under Mapping — Proof Wiki].
Item 7: Interaction With Differences
See [Proof Wiki Contributors, Image of Set Difference Under Mapping — Proof Wiki].
Item 8: Interaction With Complements
Applying Item 17 to $X\setminus U$, we have
\begin{align*} f_{!}(U^{\textsf{c}}) & = f_{!}(X\setminus U)\\ & = Y\setminus f_{*}(X\setminus (X\setminus U))\\ & = Y\setminus f_{*}(U)\\ & = f_{*}(U)^{\textsf{c}}. \end{align*}
This finishes the proof.
Item 9: Interaction With Symmetric Differences
We have
\begin{align*} f_{!}(U)\mathbin {\triangle }f_{!}(V) & = (f_{!}(U)\cup f_{!}(V))\setminus (f_{!}(U)\cap f_{!}(V))\\ & \subset (f_{!}(U)\cup f_{!}(V))\setminus (f_{!}(U\cap V))\\ & = (f_{!}(U\cup V))\setminus (f_{!}(U\cap V))\\ & \subset f_{!}((U\cup V)\setminus (U\cap V))\\ & = f_{!}(U\mathbin {\triangle }V), \end{align*}
where we have used:
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Item 2 of Proposition 4.3.12.1.2 for the first equality.
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2.
Item 6 of this proposition together with Item 1 of Proposition 4.3.10.1.2 for the first inclusion.
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Item 5 for the second equality.
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4.
Item 7 for the second inclusion.
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Item 2 of Proposition 4.3.12.1.2 for the third equality.
Since $\mathcal{P}(Y)$ is posetal, naturality is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2). This finishes the proof.
Item 10: Interaction With Internal Homs of Powersets
We have
\begin{align*} f_{!}([U,V]_{X}) & = f_{!}(U^{\textsf{c}}\cup V)\\ & = f_{!}(U^{\textsf{c}})\cup f_{!}(V)\\ & = f_{*}(U)^{\textsf{c}}\cup f_{!}(V)\\ & = [f_{*}(U),f_{!}(V)]_{Y},\end{align*}
where we have used:
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1.
Item 5 for the second equality.
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2.
Item 17 for the third equality.
Since $\mathcal{P}(Y)$ is posetal, naturality is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2). This finishes the proof.
This follows from Item 2 and
,
of
.
Item 12: Oplax Preservation of Limits
The inclusion $f_{!}(X)\subset Y$ is automatic. See [Proof Wiki Contributors, Image of Intersection Under Mapping — Proof Wiki] for the other inclusions.
Item 13: Symmetric Strict Monoidality With Respect to Unions
This follows from Item 11.
Item 14: Symmetric Oplax Monoidality With Respect to Intersections
The inclusions in the statement follow from Item 12. Since $\mathcal{P}(Y)$ is posetal, the commutativity of the diagrams in the definition of a symmetric oplax monoidal functor is automatic (Chapter 11: Categories, Item 4 of Proposition 11.2.7.1.2).
Item 15: Interaction With Coproducts
Omitted.
Omitted.
Item 17: Relation to Codirect Images
Applying Item 16 of Proposition 4.6.3.1.7 to $X\setminus U$, we have
\begin{align*} f_{*}(X\setminus U) & = B\setminus f_{!}(X\setminus (X\setminus U))\\ & = B\setminus f_{!}(U). \end{align*}
Taking complements, we then obtain
\begin{align*} f_{!}(U) & = B\setminus (B\setminus f_{!}(U)),\\ & = B\setminus f_{*}(X\setminus U), \end{align*}
which finishes the proof.