## § Left and right adjoints to inverse image

#### § The story in set

• Suppose $f: A \to B$ is a morphism of sets.
• Then there is an associated morphism $f^*: 2^B \to 2^A$, the inverse image.
• We get two associated morphisms, $\forall_f, \exists_f: 2^A \to 2^B$, which perform universal and existential quantification "relative to $f$".
• The idea is this: think of $A$ as being fibered over $B$ by $f$. Then $\forall_f(S \subseteq A)$gives the set of $b \in B$ such that the fiber of $b$ lies entirely in $A$. That is, $f^*(b) = f^{-1}(b) \subseteq A$.
• In pictures, Suppose the @ mark the subset $S$ of $A$, while the - is outside the subset. We draw $A$ as being fibered over $B \equiv \{b_1, b_2, b_3\}$.
-   @  @
-   -  @
-   -  @
|   |  |
v   v  v
b1 b2 b3

• Then, $\forall_f(A)$ will give us $b_3$, because it's only $b_3$ whose entire fiber lies in $A$.
• Dually, $\exists_f(A)$ will give us $\{ b_2, b_3 \}$, because some portion of the fiber lies in $A$.

#### § The story in general

• Suppose we have a presheaf category $\hat C$. Take a morphism $(c \xrightarrow{f} c')$

#### § The story in slice categories

• If we have $f: A \to B$ in Set, then we have $f^*: Set/B \to Set/A$, which sends a morphism $(K \xrightarrow{g} B)$ to $(K \xrightarrow{g} B \xrightarrow{f^{-1}} A)$.
• This also motivates the presheaves story, as $Set/B \simeq Set^B$.
• Recall that any morphism $K \xrightarrow{h} B \in Set/B$ can be equally seen as a morphism $b \mapsto h^{-1}(b) \in Set^B$. This is the mapping between slice and exponential.
• We can think of $(K \xrightarrow{h} B) \in Set/B$ as a collection $\{ h_b \equiv h^{-1}(b) \subseteq B \}$. This is the fibrational viewpoint.
• Then the functor $f^*(\{ h_b : b \in B\}) \equiv \{ h_{f(a)} : a \in A\}$.
• TODO