What explains the asymmetry here?

Solution 1:

The asymmetry arises because functions themselves are defined asymmetrically.

A function $f$ is defined so that:

• for all $x$ in the domain there exists a unique $y$ in the image such that $f(x)=y$

In order for the definition to be symmetric we'd also need

• for all $y$ in the image there exists a unique $x$ in the domain such that $f(x)=y$

The second point is exactly the definition of injectivity. (The first point is well-definedness.) Injectivity is equivalent to distributivity over intersections (see the other answers to this question).

The fact that all functions are well-defined corresponds with the distributivity over unions; the fact that not all functions are injective corresponds with the (lack of) distributivity over intersections: those functions which do distribute over all intersections are injective.

Solution 2:

If $y\in f[A]\cup f[B]$, then $y$ must be the image under $f$ of something in $A$ or of something in $B$, i.e., of something in $A\cup B$. If $y\in f[A]\cap f[B]$, on the other hand, we know that $y$ is the image of some $a\in A$ and the image of $b\in B$, but there’s no reason to suppose that $a$ and $b$ are the same thing, or even that there is anything at all in $A\cap B$. Of course if $f$ is injective (one-to-one), then $a$ and $b$ do have to be the same thing, and in that case we get equality of $f[A\cap B]$ and $f[A]\cap f[B]$.

I’d say that the different behavior isn’t so much a matter of some inherent asymmetry between $\cup$ and $\cap$ as it is a matter of how these Boolean operations interact with arbitrary functions.

Solution 3:

If $f:X\to Y$, then $f(A\cap B)=f(A)\cap f(B)$ for each $A,B\subseteq X$ is equivalent to $f$ being injective, hence the discrepancy in general. Again $f(A\setminus B)=f(A)\setminus f(B)\iff\text{ injection }$, so when taking the complement of your first $\bigcup$ relation, the injectivity will be needed to get to the other expression.

Solution 4:

(For fun,) I will express the asymmetry using the asymmetry between left and right adjoints in category theory. The main idea is that left adjoints commute with colimits (and unions are colimits in these categories) but not necessarily limits (and intersections are limits).

Let $f:C\to D$ be a function and let $\mathscr C$ be the category representing the partial order on the power set of $C$ given by inclusion. Similarly, construct $\mathscr D$. We have a functor $f:\mathscr C\to\mathscr D$ which assigns a set its image. The key is that it also "assigns" morphisms to morphisms, in other words it preserves order.

Define also $f^{-1}:\mathscr D\to\mathscr C$ by assigning a set its inverse image under $f$. This too is order-preserving, so it is a functor as well.

I claim $f$ is left-adjoint to $f^{-1}$. For any subset $S$ of $C$, we have both $S\subseteq f^{-1}fS$ and $ff^{-1}S\subseteq S$. To see the first, note that if $x\in S$, then $f(x)\in fS$. The second is tautological. Now, if $S\subseteq f^{-1}T$, then $f^{-1}fS\subseteq f^{-1}(ff^{-1}T)\subseteq f^{-1}T$. We just showed that the inclusion $S\subseteq f^{-1}fS$ is a universal arrow to $f^{-1}$ from $S$. The family of such arrows is clearly a natural transformation from the identity functor on $\mathscr C$ to the composite functor $f^{-1}f$. Therefore, $f\dashv f^{-1}$.

Since left adjoints commute with colimits, and union is a colimit in $\mathscr C$, we have $f(S\cup T)=f(S)\cup f(T)$. Dually, right adjoints commute with limits, but $f$ is not automatically a right adjoint. So, we cannot just say $f(S\cap T)=f(S)\cap f(T)$. In fact,

Proposition. If $f:C\to D$ induces a right adjoint $f:\mathscr C\to\mathscr D$, then $f$ is injective.
Proof. Let $g:\mathscr D\to\mathscr C$ be a functor such that if $S\subseteq D$ and $T\subseteq C$, $gS\subseteq T$ if and only if $S\subseteq fT$.

Suppose, for the sake of contradiction, that $x,y\in C$ are distinct elements such that $f(x)=f(y)$. Let $T=\{x\}$ and let $S=\{f(x)\}$. Then $gS\subseteq\{x\}$. Similarly $gS\subseteq\{y\}$ by setting $T=\{y\}$. Thus $gS=\emptyset$. In particular, $gS\subseteq C\setminus f^{-1}(f(x))$. Therefore, $S\subseteq D\setminus \{f(x)\}$, a contradiction. $\square$