If $i: U \to Z$ is an open embedding of schemes, and $\rho: Y \to Z$ is any morphism of schemes, how do we know $U \times_Z Y = \rho^{-1}(U)$? An exercise from Vakil's FOAG states to show that open embeddings are closed under fibered products.
However, we have not yet started learning about fibered product of schemes.
Vakil just gives us that

If $i: U \to Z$ is an open embedding of schemes, and $\rho: Y \to Z$ is any morphism of schemes, then $U \times_Z Y = (\rho^{-1}(U), \mathcal O_Y|{\rho^{-1}(U)})$.

We have learned about fibered product of sets, and in the category of sets, I can see that $U \times_Z Y = \rho^{-1}(U)$.
However, in the chapter of fibered product of schemes he writes that:

Warning: products of schemes aren’t products of sets. Before showing existence, here is a warning: the product of schemes isn’t a product of sets (and more generally for fibered products).


Given that I do not know about fibered products in the category of schemes, how would I show that $U \times_Z Y$ exists in the category of schemes?
 A: When it comes to Fibered products of schemes, the universal property is your friend. I encourage you to actually draw the diagrams and check that the maps are commutative, partially for your sake and partially because MSE doesn't support commutative diagrams that well.
$\rho^{-1}(U)$ comes with canonical maps $p: \rho^{-1}(U) \to U$ (obtained by restricting $\rho$) and $j: \rho^{-1}(U) \to Y$, where $j$ is just the inclusion map. Then, these maps commute (!) with the maps $i: U \to Z$ and $\rho: Y \to Z$ to form a square which we claim is cartesian.
After all, if $S$ is a scheme with $f: S \to U$, $g: S \to Y$ morphisms so that the resulting square is commutative, we can define $S \to \rho^{-1}(U)$. After all, the image of the composite morphism to $Z$ is contained in $U$ so that the image of $S$ in $Y$ is contained in $\rho^{-1}(U).$ Hence, there is a unique factorization $S \to \rho^{-1}(U) \hookrightarrow Y$ whose composite is $g: S \to Y$. This commutes with the rest of the maps (!) so that the universal property is satisfied.
EDIT: I'll add more detail about restricting scheme morphisms. Let $f: X \to Y$ be a morphism of schemes, $i: U \hookrightarrow X$ an open subscheme. We define $f|_{U}: U \to Y$ to be the composition  $f \circ i: U \hookrightarrow X \to Y$. On sheaves, we recall that the structure sheaf on $U$ is $\mathcal{O}_X|_{U}$. We see that $f|_{U}^\#: \mathcal{O}_Y \to f_*(\mathcal{O}_X|_{U})$ is the composite $$\mathcal{O}_Y(V) \stackrel{f^\#}\to \mathcal{O}_X(f^{-1}(V)) \stackrel{\rho}\to \mathcal{O}_X(f^{-1}(V) \cap U)$$ given that $\mathcal{O}_X(f^{-1}(V) \cap U) = f_*(\mathcal{O}_X|_{U})(V)$.
As an exercise, you should show that if $f: X \to Y$ is a morphism of schemes and the image of $f$ is contained in $j: V \hookrightarrow Y$ open, then there is a unique morphism of schemes $g: X \to V$ so that $f = j \circ g$. This will require you to work with the structure sheaves but it is not too difficult.
