finite direct sum of essential monomorphisms is essential Let $\mathcal{C}$ be an abelian category. A monomorphism $f:A\rightarrow B$ is called essential if $g:B\rightarrow C$ is a monomorphism whenever $gf$ is a monomorphism.
Prove that a finite family of monomorphism is essential if and only if the direct sum is essential.
The conclusion is clearly true in the category of (left) modules over a ring(cf. GTM 13 rings and categories of modules,Frank W. Anderson). 
I tried to use the embedding theorem but the  essentiality is not preserved or reflected under the embedding functor, at least I can't prove it.  
 A: Let $A \subseteq B$ and $A' \subseteq B'$ be subobjects.
If $A \oplus A' \subseteq B \oplus B'$ is essential, then $A \subseteq B$ (and likewise $A' \subseteq B'$) is essential: If a subobject $C \subseteq B$ satisfies $ C \cap A = 0$, then $C \oplus 0 \subseteq B \oplus B'$ satisfies $(C \oplus 0) \cap (A \oplus A') = 0$, which implies $C \oplus 0 = 0$, i.e. $C = 0$.
The other direction is more interesting:
Let $A \subseteq B$ and $A' \subseteq B'$ be essential. Let $C \subseteq B \oplus B'$ be a subobject satisfying $C \cap (A \oplus A') = 0$. Let $\pi_{B'} : B \oplus B' \to B'$ be the projection. There is an exact sequence
$0 \to C \cap B \to C \to \pi_{B'}(C) \to 0.$
Since $C \cap B \subseteq B$ satisfies $C \cap B \cap A = C \cap A = 0$, we see $C \cap B = 0$. Hence, $C \to \pi_{B'}(C)$ is an isomorphism. Hence, it suffices to prove $\pi_{B'}(C) = 0$. For this it suffices to prove $\pi_{B'}(C) \cap A' = 0$.
So let $a' : T \to \pi_{B'}(C) \cap A'$ be some morphism into this intersection. The induced morphism $f : T \to \pi_{B'}(C) \cong C \subseteq B \oplus B'$ consists of two components: The first one is some morphism $b : T \to B$, and the other one is $a' : T \to \pi_{B'}(C) \cap A' \subseteq A' \subseteq B'$. Since $C \cap (A \oplus A')  = 0$, we see that $\mathrm{im}(b) \cap A = 0$. Hence, we have $\mathrm{im}(b)=0$, i.e. $b=0$. But then $\mathrm{im}(f) \subseteq C \cap A' = 0$ shows $f=0$.
