Pullback of cartesian fibration preserves cartesian morphisms? Let $$P:X \rightarrow B$$ be a cartesian fibration of ordinary one category.  Let $F:C \rightarrow B$ be an arbitrary functor.  Let $$Y \xrightarrow{F^*P} C$$
be the pullback of $P$ along $F$. Then it is claimed (p16 line 1) that 

A morphism in $Y$ is cartesian wrt $F^*P$ iff its projection to $X$ is cartesian wrt $P$. 

I see if the projection to $X$ is cartesian wrt $P$, then the original morphism in $Y$ is cartesian. But I don't see the converse. May someone elaborate? 
 A: Suppose $(u,\phi) : (c,x)\to (c',x')$ is a morphism in $Y$. We want to show that $(u,\phi)$ is cartesian if and only if $\phi$ is cartesian.
Let $(v,\psi) : (c'',x'')\to (c',x')$ be another morphism in $Y$.
Then by definition, $(u,\phi)$ is cartesian if and only if for all morphisms $(v,\psi)$,
and all
$w : c'' \to c$ such that $v = uw$ there is a unique morphism $\chi : x''\to x$ such that 
(i) $\psi = \phi\chi$, and (ii) $(w,\chi)$ is a morphism of $Y$.
Note that (ii) is equivalent to saying that $F(w) = P(\chi)$.
Thus if $\phi$ is a cartesian morphism, then as you claim, we can lift $F(w)$ to produce a unique $\chi$ demonstrating that $(u,\phi)$ is cartesian.
For the converse, which is what you've asked about, we do the following.
Let $\phi' : x'''\to x'$ be a cartesian lift of $P(\phi)$.
Then by the first case $(u,\phi')$ is cartesian. Thus since both $(u,\phi)$ and $(u,\phi')$ are cartesian with the same projection and codomain, there is a unique isomorphism between them. (I.e. an isomorphism $(\alpha,\beta)$, with $\alpha : u\to u$ and $\beta : x\to x'''$).Thus since $\phi'$ is cartesian, $\phi$ must also be cartesian.
