Sometimes in mathematics, you are told about very elementary things of which you hadn't even thought.
I was well aware of some “duality” between image and preimage, but I just learned from Anatole Dedecker (who learned it from Patrick Massot) about another “duality” between preimage and some other notion. Moreover, it appears that this new notion can be used for making slightly more natural a proof in general topology!
Here, “duality” is taken in an informal meaning, the correct word is “adjunction”, in the sense of category theory, and I will try to explain that.
1. Image and preimage
So consider a map $f\colon X \to Y$ between two sets. It induces two other maps relating the sets $\mathcal P(X)$ and $\mathcal P(Y)$ of subsets of $X$ and $Y$. Note that the inclusion relation between subsets these two sets $\mathcal P(X)$ and $\mathcal P(Y)$ allows to view them as ordered sets.
First, we have the direct image operation $f_{*}$, that maps a subset $A\subseteq X$ to the subset $f_{*}(A)$ of $Y$, the set of all images $f(a)\in X$, for $a\in A$. The classical notation would be $f(A)$, but it is ambiguous in the case where a subset $A$ of $X$ is also an element of $X$, and introducing a specific notation will help to clarify some statements later on. This map $f_{*}\colon \mathcal P(X) \to \mathcal P(Y)$ is increasing: for $A$ and $A'\in\mathcal P(X)$ such that $A\subseteq A'$, one has $f_{*}(A) \subseteq f_{*}(A')$.
Then we have the preimage operation $f^{*}$, that maps a subset $B\subseteq X$ to the subset $f^{*}(B)$ of $X$ consisting of all preimages of elements of $B$, namely all $a\in A$ such that $f(a) \in B$. The classical notation is rather $f^{-1}(B)$, but it has the same ambiguity as the direct image. Bizarrely, Bourbaki found the need to invent a another notation for that one, and they put the symbol “$-1$” on top of the letter $f$. The notation $f ^{*}$ is chosen by symmetry with the direct image $f_{*}$. Again, the map $f^{*}\colon \mathcal P(Y) \to\mathcal P(X)$ is increasing: for $B$ and $B'\in\mathcal P(Y)$ such that $B\subseteq B'$, one has $f^{*}(B) \subseteq f^{*}(B')$.
Finally, there is a compatibility between these two operations $f_{*}$ and $f ^{*}$: for $A\in\mathcal P(X)$ and $B\in\mathcal P(Y)$, one has $f_{*}(A) \subseteq B$ if and only if $A \subseteq f^{*}(B)$. Indeed, both of these expressions mean that if $f(a) \in B$ for all $a\in A$. We summarize this property by saying that the operation $f_{*}$ is left adjoint to the operation $f ^{*}$, or that the operation $f^{*}$ is right adjoint to the operation $f_{*}$.
This terminology comes from category theory, in which adjunctions of functors play an important role since the paper of Daniel Kan (1958), Adjoint functors.
In our case, the categories are just the ordered sets $\mathcal P(X)$ and $\mathcal P(Y)$, with the corresponding sets as sets of objects, and where the set of arrows $A$ to $A'\in\mathcal P(X)$ is a singleton when $A\subseteq A'$, and is empty otherwise. The book of Emily Riehl (2016), Category Theory in Context, is a nice introduction to this topic, with illuminating elementary examples. The property that the operations $f_{*}$ and $f^{*}$ are increasing means that they are *functors* between these categories, and the equivalence $f_{*}(A) \subseteq B \Leftrightarrow A \subseteq f^{*}(B)$ induces the category-theoretical adjunction.
In this case, an adjunction pair is also called a Galois connection. There, the terminology comes from Galois theory, the two ordered sets are the set of subextensions of a Galois extension $K\to L$ and the set of subgroups of the Galois group $\operatorname{Gal}(L/K)$, the maps are decreasing and correspond to mapping a subextension $E$ of $L$ to the subgroup of $\operatorname{Gal}(L/E)$ of $\operatorname{Gal}(L/K)$, and a subgroup $H\subseteq \operatorname{Gal}(L/K)$ to the fixed-field $L^H$. In Galois theory, these two maps are even bijective.
2. The adjoint functor theorem
While, as MacLane wrote, “adjoint functors arise everywhere”, not every functor can be part of an adjunction. Indeed, if a functor $F$ is left adjoint to a functor $G$, then $F$ preserves colimits and $G$ preserves limits.
Category theory considers limits and colimits of arbitrary diagrams, but in the restricted setting of ordered sets, where there can be at most one arrow from one object to another, diagrams boil down to subsets of objects, limits correspond to infimums (greatest lower bound) and colimits to supremums (least upper bound), which may exist, or not, in particular ordered sets.In our even more restricted case of the set $\mathcal P(X)$ of subsets of a given set $X$, infimum corresponds to intersection, supremum to union, and we have $f_{*}(\bigcup A_i) = \bigcup f_{*}(A_i)$ for every family $(A_i)$ of subsets of $X$, and $f^{*}(\bigcap B_i) = \bigcap f^{*}(B_i)$ for every family $(B_i)$ of subsets of $Y$.
There is an abstract theorem in category theory, the “general adjoint functor theorem”, that says that these property are essentially sufficient for a functor $F$ to be a left adjoint to some functor $G$, or for a functor $G$ to be a right adjoint to some functor $G$. One has to be more careful for the actual statement, but this is the idea.
For an increasing map $G\colon T \to S$ between ordered sets $S$ and $T$, the existence of a left adjoint $F$ can be understood from: for $s\in S$ and $t\in T$, one should have $F(s)\leq t$ if and only if $s\leq G(t)$: consequently, it suffices to take for $F$ the infimum, assuming it exists, of all $t$ such that $s\leq G(t)$. Dually, the right adjoint $G$ to a functor $F$ would map $t$ to the supremum, assuming it exists, of all $s$ such that $t\leq F(s)$.
In the case of the image $f_{*}\colon \mathcal P(Y)\to \mathcal P(X)$, this rule defines the right adjoint as mapping $B \in\mathcal P(Y)$ to the union of all subsets $A\in\mathcal P(X)$ such that $f _{*}(A) \subseteq B$. This is exactly the preimage of $B$!
Conversely, in the case of the preimage $f^{*}\colon \mathcal P(Y)\to \mathcal P(X)$, this procedure defines the left adjoint as mapping $A \in\mathcal P(X)$ to the intersection of all subsets $B$ such that $A \subseteq f^{*}(B)$. Again, this is just the image $f _{*}(A)$ of $A$, but I find it slightly more difficult to prove without using that we already know this image and the already known adjunction between $f _{*}$ and $f ^{*}$.
3. The other adjunction
We have seen that preimages respect intersections. As a matter of fact, they also respect unions: $f ^{*}(\bigcup B_i)= \bigcup f ^{*}(B_i)$. Given the adjoint functor theorem, this implies that there is an increasing map $f_! \colon \mathcal P(X) \to \mathcal P(Y)$ which is a right adjoint to $f ^{*}$. What is this operation?
The adjoint functor theorem gives a way to compute it: for $A\in\mathcal P(X)$, the set $f_!(A)\in\mathcal P(Y)$ is the union of all subsets $B\in\mathcal P(Y)$ such that $f^{*}(B) \subseteq A$. It suffices to consider such sets $B$ which are singletons $\{b\}$ and we get that a point $b\in Y$ belongs to $f_!(A)$ if and only if all preimages of $b$ belong to $A$.
Here are two more ways to get a grip on this new adjunction.
Note that a point $b\in Y$ belongs to $f_{*}(A)$ if and only if there exists $a\in A$ such that $b = f (a)$, which means that there exists $a\in A$ in the preimage $f^{*}(\{b\})$, relating $f_{*}$ with the existential quantifier. Similarly, a point $b\in Y$ belongs to $f_! (A)$ if and only if for every $a\in f^{*}(\{b\})$, one has $a\in A$, relating $f_!$ with the universal quantifier.
The other way comes by taking complements: a point $b$ does not belong to $f_!(A)$ if it has a preimage that does not belong to $a$. In other words, $f_!(A) = \complement f_{*}(\complement A)$. This leads to considering the complement map from $\mathcal P(X)$ to itself as an order-reversing involution, and similarly on $\mathcal P(Y)$, and observing that they commute with preimage, in the sense that $f^{*}(\complement B) = \complement f^{*}(B)$ for all $B\subseteq Y$. Consequently, this operation transfers the left adjoint $f _{*}$ of $f ^{*}$ to a right adjoint, and conversely, which is exactly what we had observed.
4. An application in general topology
As an application, this adjunction can be used in topology to characterize open or closed maps. By definition, a map $f \colon X\to Y$ between topological spaces is open if it maps an open subset to an open subset, and it is closed if it maps a closed subset to a closed subset.
The definition of $f_!$ using complement, and the fact that a set is closed if and only if its complement is open implies the following lemma:
Lemma. — A map $f\colon X \to Y$ is closed (resp. open) if and only if for every open (resp. closed) subset $U\subseteq X$, the set $f_! (U)$ is closed (resp. open).
It also allows to give a natural proof of the classical characterization of closed maps:
Proposition. — Let $f\colon X \to Y$ be a map between topological spaces. The following properties are equivalent:
- The map $f$ is closed;
- For any subset $B$ of $Y$, the filter of neighborhoods of $f^{*}(B)$ is coarser than the preimage of the filter of neighborhoods of $B$;
- For any subset $B$ of $Y$ and any neighborhood $U$ of $f^{*}(B)$, there exists a neighborhood $V$ of $B$ such that $f^{*}(V)\subseteq U$;
- For any point $b\in Y$, the filter of neighborhoods of $f^{*}(\{b\})$ is coarser than the preimage of the filter of neighborhoods of $b$;
- For any point $b\in Y$ and any neighborhood $U$ of $f^{*}(\{b\})$, there exists a neighborhood $V$ of $b$ such that $f^{*}(V) \subseteq U$.
Given the definitions of the preimage of a filter and the comparison relation on filters,
the assertions (2) and (3) are equivalent, as well as the assertions (4) and (5).
Obviously, (3) implies (5).
Let us assume (1), that $f$ is closed, and let us prove (3). Let $B$ be a subset of $Y$ and let $U$ be a neighborhood of $f^{*}B$ in $X$. By definition, there exists an open subset $U'$ of $X$ such that $f^{*}B \subseteq U' \subseteq U$. Taking adjunction, we get $B\subseteq f_! U' \subseteq f_! U$. Since $f$ is closed, the set $f_! U'$ is open, so that $f_! U$ is a neighborhood of $B$. It remains to prove that $f^{*}f_! U\subseteq U$. To prove this inclusion, we apply the adjunction $(f^{*}, f_!)$ once more, and see that it is equivalent to the obvious inclusion $f_! U \subseteq f_! U$.
Finally, let us assume (5) and let us prove that $f$ is closed. Let $U$ be an open subset of $X$ and let us prove that $f_! U$ is open in $Y$. It suffices to prove that for every $b\in f_! U$, the set $f_! U$ is a neighborhood of $b$. By the construction of $f_!$, the set $f^{*}(\{b\}) $ is contained in $U$ so that $U$ is a neighborhood of $f^{*}(\{b\})$. Applying (5), we get a neighborhood $V$ of $b$ in $Y$ such that $f^{*}V \subseteq U$. Applying the adjunction $(f^{*}, f_!)$, we get the inclusion $V \subseteq f_! U$. In particular, $f_! U$ is a neighborhood of $b$, as was to be shown.
