I'm having trouble understanding Lemma 6 here. The part that I don't understand is where the author says that by definition $$\textrm{Im }f\ \cup\ \textrm{Im }g = \textrm{Im }(\nabla b).$$ I already saw that $\textrm{Im }f\subset\textrm{Im }(\nabla b)$ and $\textrm{Im }g\subset\textrm{Im }(\nabla b)$. But I can't see why the property defining the union of subobjects holds. Any hint? Thanks.
Edit 1. The definition of the union is the following: Suppose we have a family of subobjects $\{A_i\}$ of $A$. The union of the family is a subobject $A'$ of $A$ larger than every $A_i$ and satisfiying the following property: If we consider a morphism $f:A\rightarrow B$ and a subobject $B'$ of $B$ such that $f$ restricted to every $A_i$ may be factored through $B'$, then $f$ restricted to $A'$ may be factored through $B'$.
Edit 2. if we take the epi-mono factorizations $f=u_f p_f$, $g=u_g p_g$, then $b:\textrm{Im }f\oplus\textrm{Im }g\rightarrow B\oplus B$ Is the morphism given by $u_f\oplus u_g$ and $\nabla$ is the codiagonal. Here we are taking $f,g:A\rightarrow B$.
This is how I understand your question:
Let $u : S\to X$ and $v: T\to X$ be monomorphisms (the images of $f$ and $g$) and let $e:S\oplus T\to U $ and $w : U \to X$ be morphisms such that $me$ is the epi-mono factorisation of $\Delta (u \oplus v)$. We need to show that for any morphism $h: U \to B$ and any monomorphism $x : V\to B$ such that there exists $j : S\to V$ and $k : T\to V$ such that $xj = hei_1$ and $xk = hei_2$ (where $i_1$ and $i_2$ are the first and second biproduct inclusions) there is a morphism $h' : U \to V$ such that $h=xh'$.
If this is indeed your question, then here is an how to solve it.
(a) Show that there is a unique morphism $p : S\oplus T \to V$ such that $pi_1=j$ and $pi_2=k$.
(b) Show that $ve=xp$.
(c) Using (b) show that $p\text{ker}(e)=0$.
(d) Using that epis are normal epis (i.e. cokernels of some morphism and hence of their kernels) show that there is a unique morphism $h' : U\to V$ such that $h'e=p$.
(e) Show that $h = xh'$.