# A counterexample for the pullback $id_{U}^{*}$ not be an identity

Definition $1$.- Let $\mathcal{P_F:F}\rightarrow\mathcal{C}$ be a functor. Given an object $U$ of $\mathcal{C}$, the fiber of $\mathcal{F}$ over $U$, denoted by $\mathcal{F}(U)$, is the subcategory of $\mathcal{F}$ whose objects are the objects $\zeta$ of $\mathcal{F}$ with $\mathcal{P_F}(\zeta)=U$ and whose arrows are the arrows $\phi$ in $\mathcal{F}$ with $\mathcal{P_F}(\phi)=id_U$.

Definition $2$.-Let $\mathcal{F}$ be a category fibered over $\mathcal{C}$ and $f:U\rightarrow V$ an arrow in $\mathcal{C}$. For each object $\eta$ over $V$, we choose a pullback $\phi_{\eta}:f^*\eta\rightarrow \eta$ o $\eta$ to $U$ and defines a functor $f^*:\mathcal{F}(V)\rightarrow\mathcal{F}(U)$ by sending each object $\eta$ of $\mathcal{F}(V)$ to $f^*\eta$, and each arrow $\beta:\eta\rightarrow\eta'$ of $\mathcal{F}(V)$ to the unique arrow $f^*\beta:f^*\eta\rightarrow f^*\eta'$ in $\mathcal{F}(U)$ making the diagram

$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} f^*\eta & \ra{} & \eta \\ \da{f^*\beta} & & \da{\beta} \\ f^*\eta' & \ras{} & \eta' \\ \end{array}$$

commute.

Definition $3$.- A cleavage of a fibered category $\mathcal{F}\rightarrow\mathcal{C}$ consists of a class $K$ of cartesian arrows in $\mathcal{F}$ such that for each arrow $f:U\rightarrow V$ in $\mathcal{C}$ and each object $\eta$ in $\mathcal{F}(V)$ there exists unique arrow in $K$ with target $\eta$ mapping to $f$ in $\mathcal{C}$.

Now, given a fibered category $\mathcal{F}\rightarrow\mathcal{C}$ with a cleavage, define a map that sends $U\in\mathcal{C}$ to the subcategory $\mathcal{F}(U)$, and a morphism $f:U\rightarrow V$ of $\mathcal{C}$ is mapped to $f^*:\mathcal{F}(V)\rightarrow \mathcal{F}(U)$, this is not a functor from $\mathcal{C}$ to $CAT$ the category of categories because, for example, all pullbacks $id_{U}^{*}:\mathcal{F}(U)\rightarrow\mathcal{F}(U)$ are not identities.

For this last phrase, is there an example for the pullback $id_{U}^{*}$ not be an identity?

Appreciate any suggestion!

The pullback is essentially never an identity. For example, in a category $C$ with pullbacks, evaluation at the codomain gives a fibration of the arrow category over $C$. Here the pullback along the identity of $U$ sends an arrow $f:T\to U$ to the pullback $T \times_U U\to U$. Of course, $T\times_U U$ is canonically isomorphic to $T$, but the two objects should not be expected to be equal. For instance, if $C$ is the category of sets, then $T\times_U U$ is usually defined as the set of ordered pairs $(t,f(t))$, which is not equal to $T.$
Any isomorphism $\varphi$ in $\mathcal F$ is a cartesian arrow: given another map $\psi$ with same target as $\varphi$, the unique arrow filling the triangle will just be $\varphi^{-1}\psi$.
So in a cleavage $K$ of the fibered category $\mathcal F \to \mathcal C$, the chosen cartesian morphism $\eta' \to \eta$ above $\mathrm{id}_U$ can be any isomorphism of $\mathcal F(U)$. That is why $\mathrm{id}_U^\ast$ is not necessarily the identity (even if it is naturally isomorphic to the identity). So in the end, the mapping $U \mapsto \mathcal F(U)$ is a pseudo functor $\mathcal C\to \mathsf{CAT}$ rather than a functor.
Remark that the previous argument shows that if each $\mathcal F(U)$ has no non trivial isomorphism, then $\mathrm{id}_U^\ast$ must be the identity functor. This is in particular the case when each fiber is a set, and you recover the usual correspondence between presheaves and category of elements.