Do fibrations preserve limits? If I understand correctly, the fourth property here says the following.

Let $ \begin{smallmatrix}\mathsf C\\ \downarrow\\ \mathsf B
 \end{smallmatrix}\varphi$ be a cloven fibration and let $D:\mathsf
 J\to \mathsf C$ be a diagram. Let $(L,\lambda)=\varprojlim
 (\varphi\circ D)$ (suppose it exists) and define $\tilde D:\mathsf J\to \mathsf
 \varphi^{-1}(L)$ on objects by $\tilde D(J)=\lambda_J^\ast (DJ)$. Then
   $\varprojlim \tilde D=\varprojlim D$.

Since $\varprojlim \tilde D$ lies in $\varphi ^{-1}(L)$ we have $\varphi (\varprojlim \tilde D)=L$. But $\varprojlim \tilde D=\varprojlim D $ and so $$\varphi(\varprojlim D)\cong \varprojlim (\varphi\circ D)$$ i.e $\varphi$ preserves limits.
Is this correct? Do fibrations preserve limits? If not, what am I missing here?
 A: The boxed claim seems to be false. In the particular case when $\mathsf J$ is the empty category, it says that an object of the total category is terminal if and only if it is above a terminal object of the base and it is terminal in its own fiber. Below is a counter-example.
Let $C : \mathbf 2^{\rm op} \to \mathsf{Cat}$ be the (pseudo) functor from the opposite of the walking arrow $\mathbf 2 = \{ d \to r \}$ to the 2-category of small categories defined on object as:
$$ d \mapsto \mathbf 2, \quad r \mapsto \mathbf 1$$
(where $1$ is the terminal category) and defined on the unique non identity morphism $d \to r$ as the functor $1 \to 2$ that picks the object $d$ out. The Grothendieck construction $\int C \to \mathbf 2$ is a fibration whose domain can be described as a walking span $x \gets y \to z$, the left leg being mapped to $\mathrm{id}_d$ and the right one to $d\to r$. Clearly $z$ is terminal in its fiber (it is alone!) and above the terminal object $r$ of $\mathbf 2$. But it isn't terminal in $\int C$ because there is no arrow $x \to z$.
