If there is a map $Y \rightarrow Z$, then there is a map $X_1 \times_Y X_2 \rightarrow X_1 \times_Z X_2$

Question

I'm trying to solve this exercise from Ravi Vakil's notes on AG, which is labeled as trivial there but doesn't seem to be as such. It is assumed that both $X_1 \times_Y X_2$ and $X_1 \times_Z X_2$ exist for given maps $\alpha_{1,2}: X_1,X_2 \rightarrow Y$, $\beta_{1,2}:X_1,X_2 \rightarrow Z$. Also label $\pi^Y_{1,2}:X_1 \times_Y X_2 \rightarrow X_1,X_2$, $~\pi^Z_{1,2}:X_1 \times_Z X_2 \rightarrow X_1,X_2$, $~f:Y \rightarrow Z$. Sorry in advance for the absence of a diagram, but I couldn't think of anything other than something like here in the first answer (without the bottom right half) and it seems wrong to me for the reasons I describe below.

My first solution attempt was to notice, that $\alpha_1 \circ \pi^Y_1 = \alpha_2 \circ \pi^Y_2$ and therefore $f \circ \alpha_1 \circ \pi^Y_1 = f \circ \alpha_2 \circ \pi^Y_2$, so we have two maps $X \times_Y X \xrightarrow{\pi^Y_{1,2}} X_1,X_2 \xrightarrow{f\circ \alpha_{1,2}} Z$ that commute, therefore by universal property of $X \times_Z X$ we have a unique map $h: X \times_Y X \rightarrow X \times_Z X$.

However, by definition of fibered product, $X \times_Y X \xrightarrow{\pi^Y_{1,2}} X_1,X_2 \xrightarrow{\beta_{1,2}} Z$ must be valid for the universal property to be appliable. In words, maps to $X_1, X_2$ must commute in $Z$ through the maps given in $X\times_ZX$, not just any maps. Is my understanding correct? If yes, might it be the case that $f \circ \alpha_{1,2} = \beta_{1,2}$ for some reason? Or did I go in a completely wrong direction?

EDIT: I reread the exercise and it might actually be assumed that $f \circ \alpha_{1,2} = \beta_{1,2}$, or better to say, $\beta_{1,2}$ are defined as such. Here's the exercise: So sorry to everyone who tried with me to solve non-existent and unsolvable exercise. Maybe someone can post a counterexample to the construction in question? Otherwise the question can be closed.

Answer 1

The point of the exercise may have been to reinforce that a morphism to a pullback object is a pair of maps: $$ (X → A \times_C B) \quad≅\quad (X → A) ×_{(X → C)} (X → B) $$

With this, the posed problem is $``$trivial$``$: \begin{align*} X₁ ×_Y X₂ & → X₁ ×_Z X₂ & \\ \cong (X₁ ×_Y X₂ → X₁) & ×_{X₁ ×_Y X₂ → Z} (X₁ ×_Y X₂ → X₂) \end{align*}

At this point, the projections $πᵢ : X₁ ×_Y X₂ → X_i$ are two such solutions $x, y$ to the equation $f ∘ α₁ ∘ x = f ∘ α₂ ∘ y$ where both sides are of type $X₁ ×_Y X₂ → Z$, provided $f ∘ αᵢ = βᵢ$.