Generalization of the fact that pre-images of a function preserve more set operations than images

Question

Given a function $f: A \rightarrow B$ images of subsets of $A$ preserve only inclusion and union of sets whereas pre-images of subsets of $B$ are better behaved, so to speak, and preserve (in addition to the above) intersections and differences of sets as well. I have both the intuition and the proofs for the this statement.

I am trying to understand what's the deeper truth behind this.

I think that given a set $C$, a collection of sets $\mathcal{D}$, and a function $g:C\rightarrow\mathcal{D}$ with the property that for every $c1,c2 \in C$ we have that $c1\neq c2 \implies g(c1) \cap g(c2) = \emptyset$ then $g$ preserves all set operations on subsets of $C$. The pre-image is clearly such a function from individual elements of (or singleton sets in) $B$ to the powerset of $A$ so it follows that the pre-image applied to arbitrary subsets of $B$ (not necessarily singletons) preserves set operations.

1. Is this correct?

2. Is this the deepest / most general intuition one can have as to why the pre-image of a function preserves more set operations than the image?

Answer 1

For question 1., assuming that $g$ is extended to the powerset of $C$ via $$\tilde{g}(A)=\bigcup_{c\in A}g(c),$$ the answer is yes - this function will preserve all operations. This is most easily seen by noting that this $\tilde{g}$ is in fact almost a preimage function in disguise - namely of the partial function $f:D\to C$ defined by $$f(d)=c\iff d\in g(c).$$

In that sense, your definition gives essentially nothing new, but rather reformulates a key property of the preimage.

Your second question is of course a lot more open and there are surely different ways to think about this. I can offer up one perspective that may be of interest if you are comfortable with basic category theory. When interpreting this innocuous situation in a category theoretic way, the behaviour of the image and preimage is a hands-on instance of the fact that left adjoint functors preserve colimits and right adjoints preserve limits:

The starting observation is that any partially ordered set $(P,\leq)$ can be regarded as a category if we regard the points of $P$ as objects and give a single morphism from $p$ to $q$ if and only if $p\leq q$ (and nothing otherwise).

In particular, any powerset $\mathcal{P}(A)$ is a category in this way, when we consider it with the natural ordering of set inclusion.

The upshot of this setup is that the preimage and image operations take on an enhanced meaning: Any set function $f:A\to B$ gives rise to two functors $$f_*:\mathcal{P}(A)\to \mathcal{P}(B), C\mapsto f(C)$$ and $$f^{-1}:\mathcal{P}(B)\to \mathcal{P}(A),D\mapsto f^{-1}(D).$$ (You should check that these are really functors if you are unsure.) Not only that, but these functors are even adjoint, which in this situation boils down to the fact that $$f(C)\subseteq D \iff C\subseteq f^{-1}(D)$$ (again, check this if you are unsure).

And this relation, finally, is the "reason" that image and preimage have some of their nice properties: As a left adjoint, the image $f_*$ preserves colimits (which are unions in this context), while the preimage, as a right adjoint, preserves limits (which are intersections).

But, as you point out yourself, the preimage also preserves unions, while the image does not preserve intersections, so what is the difference? Well, the reason for that is the existence of a third functor $f^+:\mathcal{P}(A)\to \mathcal{P}(B)$ that I did not mention before, namely $$f^+:\mathcal{P}(A)\to \mathcal{P}(B), C\mapsto \{d:f^{-1}(\{d\})\subseteq C\}.$$ This third functor turns out to be a right adjoint of $f^{-1}$ by virtue of $$f^{-1}(D)\subseteq C \iff D\subseteq f^+(C).$$

In particular, this explains why $f^{-1}$ has both properties - as it has adjoints on both sides, it must preserve both limits and colimits (which are, again, nothing but intersections and unions in this context).

You can then check directly that for a general function $f$, this adjoint chain stops here - in particular there is no category theoretic "reason" that $f_*$ should preserve intersections, and indeed it does not.

Answer 2

Your intuition is correct and I will just make it more formal by first giving the appropriate definitions and then by summarizing, in a second part, the behaviour of functions and relations with respect to set operations. The proofs are omitted, but please ask me if you need them.

Definitions on relations.

Let $E$ and $F$ be two sets. A relation on $E$ and $F$ is a subset of $E \times F$. A relation $\tau$ can also be viewed as a function from $E$ to ${\cal P}(F)$ by setting, for each $x \in E$, $$ \tau(x) = \{ y \in F \mid (x, y) \in \tau \} $$ By abuse of language, we say that $\tau$ is a relation from $E$ to $F$.

The inverse of a relation $\tau \subseteq E \times F$ is the relation $\tau^{-1} \subseteq F \times E$ defined by $$ \tau^{-1} = \{ (y, x) \in F \times E \mid (x, y) \in \tau \} $$ It can be also viewed as a function from $F$ to ${\cal P}(E)$ defined by $$ \tau^{-1}(y) = \{x \in E \mid y \in \tau(x) \} $$ A relation from $E$ to $F$ can be extended to a function from ${\cal P}(E)$ to ${\cal P}(F)$ by setting, for each subset $X$ of $E$, $$ \tau(X) = \bigcup_{x \in X}\tau(x) = \{ y \in F \mid \text{for some $x \in X$, $(x, y) \in \tau$} \} $$ If $Y$ is a subset of $F$, we then have \begin{align*} \tau^{-1}(Y) &= \bigcup_{y \in Y} \tau^{-1}(y) = \{ x \in E \mid \text{there exists $y \in Y$ such that $y \in \tau(x)$} \}\\ &= \{ x \in E \mid \tau(x) \cap Y \neq \emptyset \} \end{align*} The notion of an injective function can be extended to relations as follows. A relation $\tau: E \to F$ is called injective if, for every $u, v \in E$, $\tau(u) \cap \tau(v) \neq \emptyset$ implies $u = v$. This is exactly your definition.

Proposition. Let $\tau$ be a relation from $E$ to $F$. The following conditions are equivalent:
$\quad (1)$ $\tau$ is injective,
$\quad (2)$ its inverse is a partial function,
$\quad (3)$ for all $X, Y \subseteq E$, $X \cap Y = \emptyset$ implies $\tau(X) \cap \tau(Y) = \emptyset$,
$\quad (4)$ $\tau^{-1} \circ \tau = Id_{\text{Dom}(\tau)}$,
$\quad (5)$ $\tau^{-1} \circ \tau \subseteq Id_E$

Set operations.

Let $\tau:E \to F$ be a relation. Then, for every $X, Y \subseteq E$,
$$ \tau(X \cup Y) = \tau(X) \cup \tau(Y) $$ Furthermore, if $\tau$ is injective, then $$ \tau(X \cup Y) = \tau(X) \cup \tau(Y) \quad \tau(X \cap Y) = \tau(X) \cap \tau(Y) \quad \tau(X - Y) = \tau(X) - \tau(Y) $$ Let $\varphi:E \to F$ be a function. Then, for every $X, Y \subseteq F$, $$ \varphi^{-1}(X \cup Y) = \varphi^{-1}(X) \cup \varphi^{-1}(Y) \quad \varphi^{-1}(X \cap Y) = \varphi^{-1}(X) \cap \varphi^{-1}(Y) \quad \varphi^{-1}(X^c) = (\varphi^{-1}(X))^c $$ Let me conclude this post with a less well-known, but useful property.
Let $\varphi:E \to F$ be a partial function. Then for every $X \subseteq E$ and $Y \subseteq F$, $$ \varphi(X) \cap Y = \varphi\bigl(X \cap \varphi^{-1}(Y)\bigr) $$