I am trying to figure out the following exercise from Vakil's notes: If the two squares in the following commutative diagram are Cartesian diagrams, then the "outside rectangle" (involving U,V,Y, and Z) is also a Cartesian diagram.
$\require{AMScd}$ \begin{CD} U @>>> V\\ @V V V @VV V\\ W @>>> X \\ @V V V @VV V\\ Y @>>> Z \end{CD}
So we want to show that for any object $R$ such that there exist morphisms $R\rightarrow Y$ and $R\rightarrow V$ making the outside square commute, then there exists a unique morphism $R\rightarrow U$.
I know this is probably just some diagram chase with unwinding of definitions, but I'm having trouble sorting through it. Let $\alpha:Y\rightarrow Z$, $\beta:X\rightarrow Z$ and $\beta':V\rightarrow X$ be labels for morphisms given above, and suppose we have an object $R$ with maps $P_{RY}:R\rightarrow Y$ and $P_{RV}:R\rightarrow V$ such that $\alpha\circ P_{RY}=\beta\circ\beta'\circ P_{RY}$.
I am not sure how to proceed. It seems there is some ambiguity in how to choose the map from $W$ to $X$ since it involves both squares. What should I do?
Since the lower square is cartesian, you will find first a map from $R$ to $W$. This map gives you a map from $R$ to $U$ using that the upper square is cartesian. So you have the existence of the map. For the uniqueness, two maps from $R$ to $U$ making commutative the triangles of rectangle would give two maps from $R$ to $W$ make commutative the triangles of the lower square, so they would be equal (from $R$ to $W$). So the two maps from $R$ to $U$ would make commutative the upper triangles and would be equal sinthe the upper square is cartesian.