What are pullbacks of finite-coproduct injections along arbitrary morphisms?

I am studying a definition of an extensive category:

An extensive category is a category $E$ with finite coproducts such that pullbacks of finite-coproduct injections along arbitrary morphisms exist and finite coproducts are disjoint and stable under pullback.

I would like to verify my understanding of the part "pullbacks of finite-coproduct injections along arbitrary morphisms exist": Let $f_A:A \to C$ and $f_B:B \to C$ be arbitrary morphisms.

1. Does that part imply that there exists a pullback $P$ with some morphisms $p_A:P \to A$ and $p_B:P \to B$ such that $f_A \circ p_A = f_B \circ p_B$?

2. Are these morphisms $p_A$ and $p_B$ (called) finite-coproduct injections? But then how does their finite-coproduct relate to the pullback $P$?

To be honest, I think I am lost. If I could see the words transformed to the symbols, it would remove ambiguity and uncertainty for me.

• No. A coproduct injection is a morphism that appears as part of a coproduct cocone. These morphisms can be pulled back along arbitrary morphisms. I do not think expressing this in symbols will be helpful if you do not know the definitions of the phrases used here. – Zhen Lin Jun 29 '16 at 0:40
• @ZhenLin What does it mean to pullback some morphisms along some other morphisms? How many morphisms are involved in such a construction? 2 or 4? Perhaps, stating generalities in a singular whenever possible would help to eliminate my confusion. – Dávid Natingga Jun 29 '16 at 0:42

i) The pullback of the morphism $f_A:A \to C$ along the morphism $f_B:B \to C$ is the morphism $p_B:P \to B$ where $\langle P, p_A, p_B \rangle$ is the pullback, i.e. $f_A \circ p_A = f_B \circ p_B$ where $P$ is universal.
ii) Then the sentence "pullbacks of finite-coproduct injections along arbitrary morphisms exist" means the following: Let $i_A:A \to A \coprod B$ be a coproduct injection and let $f:D \to A \coprod B$ be an arbitrary morphism. Then the pair of the arrows $\langle i_A, f \rangle$ has a pullback.