$J$-shaped diagram in $\mathcal{C}$ is a functor $D: J\rightarrow \mathcal{C}$. Now, a limit of $D$ is a universal cone, which is an object $Y$ together with morphisms $f_i:Y\rightarrow D(X_i)$ for all objects $X_i \in \operatorname{obj}(J)$ (also all triangles must commute etc.)
Now, in the definition of pullback, we have three objects $A,B,C$ and morphisms $\pi_B : B\rightarrow C$ and $\pi_A : A\rightarrow C$. Pullback is then defined as a object $P$ together with $p_A : P\rightarrow A, p_B : P\rightarrow B$ satisfying the universal property (i.e. for any other $Y$ together with morphisms $q_A,q_B$ there is unique $\varphi$ s.t. $q_A = p_A \circ \varphi, q_B = p_B \circ \varphi$).
Sometimes this is also regarded as a limit of the pullback diagram. But hold on, shouldn't the limit have morphisms to all objects ? I.e. also to the object $C$? What is going on here?
I think simillar problem arises with (co)-equalizers. There we have two parallel morphisms $f,g : A\rightarrow B$ and equalizer is object $E$ with morphism $e:E\rightarrow A$ but nothing is said about morphism to the object $B$. Same question arises.
EDIT: According to the comment the third morphism to the object $C$ is implicitly given by $\pi_A \circ p_A = \pi_B \circ p_B$. But consider additional cone $P$ together with morphisms $p_A, p_B$ and some $p_C$ different from $\pi_A \circ p_A$. The universal property assures existence of unique $\varphi$ s.t. $p_A = \varphi \circ p_A$ and $p_B = \varphi \circ p_B$ and $p_C = \varphi \circ \pi_A p_A$. The first two equations force $\varphi$ to be the identity, but $p_C$ was chosen to be different from $\pi_A p_A$. This is weird.