We now consider the exact sequence:
$(A \cap B, *) \xrightarrow{\Delta} (A \vee B, *) \xrightarrow{\pi} (A \cup B, *)$

with the maps as:
$\begin{aligned}
&ab \in A \cap B \xmapsto{\Delta} (0, ab), (1, ab) \in A \vee B \\
&(0, a) \in A \vee B \xmapsto{\pi}
\begin{cases}
* & \text{if } a \in B \\
a &\text{otherwise} \\
\end{cases} \\
&(1, b) \in A \vee B \xmapsto{\pi}
\begin{cases}
* & \text{if } b \in A \\
b &\text{otherwise} \\
\end{cases} \\
\end{aligned}$

- 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:
$\begin{aligned}
(A \cap B, *) \xrightarrow{\Delta} (A \times B, (*, *)) \xrightarrow{\pi} (A \cup B, *)
\end{aligned}$

with the maps as:
$\begin{aligned}
&ab \in A \cap B \xmapsto{\Delta} (ab, ab) \in A \times B \\
&(a, b) \in A \times B \xmapsto{\pi}
\begin{cases}
* & \text{if } a = b \\
a, b &\text{otherwise} \\
\end{cases} \\
\end{aligned}$

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 ].