The intersection of $u_{A} : A \longrightarrow A + B$ and $u_{B} : B \longrightarrow A+B$ is zero.

Question

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.

Answer 1

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.

Answer 2

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 ...