The intersection of $u_{A} : A \longrightarrow A + B$ and $u_{B} : B \longrightarrow A+B$ is zero. I am trying to show that the intersection of $u_{A}:A \longrightarrow A+B$ and $u_{B}:B \longrightarrow A+B$ is the zero map. Here, the $u_{A}$ and $u_{B}$ are the embedding maps into the coproduct of $A$ and $B$. 
I know how to prove that intersections exist in Abelian categories, and according to Freyd the proof of the above question follows from the construction of intersections.
I can't seem to make it work.
Any help is appreciated.
Thanks.
 A: Let $f : C \to A$ and $g : C \to B$ be morphisms such that $u_A \circ f = u_B \circ g$. Let $p_A : A \oplus B \to A$ and $p_B : A \oplus B \to B$ be the canonical projections. Then $p_A \circ u_A = \mathrm{id}_A$, $p_B \circ u_B = \mathrm{id}_B$, $p_A \circ u_B = 0$, and $p_B \circ u_A = 0$; hence:
$$f = p_A \circ u_A \circ f = p_A \circ u_B \circ g = 0$$
$$g = p_B \circ u_B \circ g = p_B \circ u_A \circ f = 0$$
Therefore the intersection of $u_A$ and $u_B$ must be zero.
A: This is Prop 2.33 of Freyd's Abelian Categories. The construction of the intersection of:$\newcommand{\Ker}{\operatorname{Ker}}$
$u_A:A\rightarrow A+B$ and $u_B:B\rightarrow A+B$ is as follows (Th 2.13):
Let $A+B\rightarrow F = \text{Cok}(u_A)$
and $A_{12}\rightarrow B = \Ker(B\rightarrow A+B\rightarrow F)$
then $A_{12}\rightarrow A+B$ is the intersection of $u_A$ and $u_B$.
So in Prop 2.33:
$A+B\rightarrow F = \text{Cok}(u_A) = (0/1)$ by the last statement of prop 2.22
and let
$A_{12}\rightarrow B = \Ker(B\rightarrow A+B\rightarrow B) = \Ker((0/1)u_A) = \ker(B\rightarrow B) = \Ker(id_B) = 0\rightarrow B$
So the intersection is:
$A_{12}\rightarrow A+B = 0\rightarrow A+B$
I really do not know how to format my answer ...
