Does the intersection of sets have a categorical interpretation?

Question

My question is the title, really. I am wondering if the intersection of sets can be seen as a categorical construction on the objects of $\mathbf{Set}$.

Answer 1

If $A, B \subseteq C$, $i_1$ is the inclusion from $A$ to $C$ and $i_2$ is the inclusion from $B$ to $C$, then consider the pullback of $i_1$ and $i_2$.

Answer 2

$\require{AMScd}$ The more useful category to consider this in is not $\text{Set}$ but the subcategory in which we only consider injective morphisms (inclusions). In that case $A\cap B$ and $A \cup B$, for $A,B \subseteq C$, fit in these pullback and pushout diagrams:

$$ \begin{CD} A\cap B @>>> A \\ @VVV & @VVV \\B @>>> C \end{CD}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \begin{CD} A \cap B @>>> A \\ @VVV & @VVV \\B @>>> A \cup B \end{CD}$$

In $\text{Set}$ one gets the same answers in the special case of injective maps, but pullbacks and pushouts in $\text{Set}$ are not in gneral intersections and unions. To characterize $A\cap B$ and $A\cup B$ as such we need to restrict to the subcategory with injective maps.

The fact that $A\cap B$ and $A\cup B$ are pullback and pushouts are essential to the way Grothendieck topologies generalize topological spaces. One replaces the open sets of a topological space, considered as inclusion maps, with other classes of morphisms with similar formal properties. Intersections of open sets then get replaced with taking fibre products (pullbacks) of morphisms.

Answer 3

No, the intersection of two "isolated" sets $A,B$ doesn't have any categorical interpretation. Because for any reasonable meaning of this, we would like to have $A \cap B \cong A' \cap B'$ if $A \cong A'$ and $B \cong B'$. But this is clearly wrong (take $A=B=A'=\{1\}$ and $B'=\{0\}$).

Thus, from the perspective of category theory, the set-theoretic operation $\cap$ doesn't make much sense. What is $\pi \cap \mathbb{R}$ supposed to be? However, it is meaningful to take the intersection with respect to two (injective) maps $A \to C$ and $B \to C$. Namely, then the pullback $A \times_C B$ is the desired intersection. For more on this, see also math.SE/295800 and math.SE/704593 and math.SE/866127.