This was a shower thought. I don't even if these form an abelian category. Let's assume we have pointed sets, where every set has a distinguished element $*$. $p$ will be analogous to the zero of an abelian group. We will also allow multi-functions, where a function can have multiple outputs. Now let's consider two sets, $A, B$ along with their 'smash union' $A \vee B$ where we take the disjoint union of $A, B$ with a smashed $*$. To be very formal:

We now consider the exact sequence:

with the maps as:

• We note that $\Delta$ is a multi-function, because it produces as output both $(0, ab)$ and $(1, ab)$.
• $\ker(\pi) = \pi^{-1}(*) = \{ (0, a) : a \in B \} \cup \{ (1, b) : b \in A \}$
• Since it's tagged $(0, a)$, we know that $a \in A$. Similarly, we know that $b \in B$.
• Hence, write $\ker(\pi) = \{ (0, ab), (1, ab) : ab \in A \cap B \} = im(\Delta)$

This exact sequence also naturally motivates one to consider $A \cup B - A \cap B = A \Delta B$, the symmetric difference. It also gives the nice counting formula $|A \vee B| = |A \cap B| + |A \cup B|$, also known as inclusion-exclusion.

I wonder if it's possible to recover incidence algebraic derivations from this formuation?

§ Variation on the theme: direct product

This version seems wrong to me, but I can't tell what's wrong. Writing it down:

with the maps as:

One can see that:

• $A \cap B \xrightarrow{\Delta} A \times B$ is injective
• $A \cap B \xrightarrow{\pi} A \cup B$ is surjective
• $ker(\pi) = \pi^{-1}(*) = \{ (a, b) : a \in A, b \in B, a = b \} = im(\Delta)$

Note that to get the last equivalence, we do not consider elements like $\pi(a, *) = a, *$ to be a pre-image of $*$, because they don't exact ly map into $*$ [pun intended ].