## § Exact sequence of pointed sets

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:
$A \vee B = \{0\} \times (A - \{ * \}) \cup \{1\}\times (B - \{ * \}) \cup \{ * \}$
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 ].