I am trying to learn some concepts of category theory and there is something I don't understand about categorical limits. I found these notes online with the following definitions (please see the notes for the diagrams).
Definition 3.11 : Given a functor $F : J\to C$, a cone of $F$ is an object $N\in C$ together with morphism $\psi_X : N\to F(X)$ for every $X\in J$ such that for every morphism $f : X\to Y \in J$, the triangle commutes, i.e. $Ff\circ \psi_X =\psi_Y$.
Definition 3.12: A limit of a functor $F : J\to C$ is a universal cone $L, \phi_X$. That is, for every cone $N$, $\psi_X$ of $F$, there is a unique morphism $u : N\to L$ such that $\phi_X = u\circ \psi_X$ for every $X \in J$. The limit object may be written $\lim\limits_{i\in J}F(i)$.
Definition 4.1. A pullback is a limit of a diagram of the following form : $A\rightarrow C\leftarrow B$. That is, a pullback is an object $D$ with morphisms $p_1 : D\to A$ and $p_2 : D\to B$ which makes the square commute and are universal, i.e. for every other object $Q$ with morophisms $q_1 : Q\to A$ and $q_2 : Q\to B$, there is a unique morphism $u : Q\to D$ which makes the diagram commute. THe pullback, interpreted as the object $D$, may be written $A\times_C B$.
Question : Applying diagram of definition 3.11, to the pullback, shouldn't we have arrows $\psi_A$, $\psi_B$ and $\psi_C$ going from $Q$(which plays the role of $N$?) to $A, B$ and $C$ respectively? Why is the arrow from $Q$ to $C$ missing? I consulted other references (e.g. Category for the working mathematicians) but it seems to be the same. Can somebody explain to me what I got wrong?
Leaving out $\psi_C$ and adding the requirement that the square commutes amounts to the same thing as including $\psi_C$ and requiring the various triangles to commute ($Ff \circ \psi_X = \psi_Y$). Since having fewer morphisms is simpler, $\psi_C$ is often left out when specifically defining pullbacks. Using just $\psi_A$ and $\psi_B$ and the square commutativity, we can uniquely reconstruct $\psi_C$.
Let $f : A \to C$ and $g : B \to C$ be the morphisms we want to take the pullback of.
For one direction, suppose that $N$ is a cone on this diagram with the morphisms $\psi_A$, $\psi_B$ and $\psi_C$ from $N$ to $A$, $B$ and $C$ respectively. The commutativity requirement says that $f \circ \psi_A = \psi_C$ and $g \circ \psi_B = \psi_C$. Then $f \circ \psi_A = \psi_C = g \circ \psi_B$, so the square commutes.
For the other direction, suppose we only have $\psi_A$ and $\psi_B$ with the condition that $f \circ \psi_A = g \circ \psi_B$. We're forced to define $\psi_C$ as either $f \circ \psi_A$ or $g \circ \psi_B$ since the triangle commutativity requirement says that $\psi_C$ is equal to both of these. Fortunately, the square commutativity requirement says that these two choices are equal. So there's a unique definition of $\psi_C$ from $\psi_A$ and $\psi_B$ plus the square commutativity requirement.