A definition of Grothendieck fibrations without cartesian morphisms. Is this possible? I am thinking of a definition of (cartesian) fibration using the classical example of fibrations: Commutative squares in a category with pullbacks C fibred over C.
Definition: Given a functor $p:X \to C$, we say $p$ is a fibration if given any object $x \in X$ and a morphism $\alpha: c \to p(x)$, there exists a "universal lift" $\beta: x' \to x$ in $X$ such that $p(\beta) = \alpha$ with the property that for any other lift $\gamma: x'' \to x$ in $X$ (i.e. $p(\gamma) = \alpha$), the morphism $\gamma$ factors through a unique morphism $\theta: x'' \to x'$ (i.e. $\beta \circ \theta = \gamma$).
It seems to me that we should be able to define it using "universal lifts".
If this definition works, then morphisms in $X$ that arise as universal lifts will be cartesian morphisms.
I am not able to show the above definition is equivalent to the notion of Grothendieck fibrations. Maybe it is not! So please prove me wrong.
If there are references along this direction, I appreciate that as well.

If you require that the lift factors through a unique morphism $\theta$ such that $p(\theta)$ is the identity, then you would have asserted that every morphism has a unique weak Cartesian lift. According to Exercise 1.1.6 in Jacobs' book Categorical Logic and Type Theory, the functor is a fibration  if and only if every morphism has a unique weak Cartesian lift and weak Cartesian morphisms are closed under composition.
