TITLE: What explains the asymmetry here?
QUESTION [13 upvotes]: The image operation distributes over unions:
$$f(A \cup B) = f(A) \cup f(B)$$
but about intersections we can only say that
$$f(A \cap B) \subseteq f(A) \cap f(B)$$
unless $f$ is injective. Where does this asymmetry arise from?

REPLY [2 votes]: (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. $\square$