I have seen this used as an argument in my textbook:
Set-up:
Assume $\mathcal{C}$ is a full subcategory of an Abelian category $\mathcal{A}$. Let $f: A \rightarrow B$ be a morphism in $\mathcal{C}$.
If $\text{ker }f$ lies in $\mathcal{C}$ then $\text{ker }f$ is the same in $\mathcal{A}$.
(They also assumed in the textbook that the zero object in $\mathcal{A}$ lies in $\mathcal{C}$ and the direct sum of $A,B \in \text{Obj} \mathcal{C}$ lies in $\mathcal{C}$ and the $\text{coker} f$ lies in $\mathcal{C}$.)
Is this true? And how do I prove it?
I think my problem is my understanding of the definition of the kernel. In the textbook the kernel of $f$ in $\mathcal{C}$ should have the property: Let $i:K\rightarrow A$ be the kernel of $f$ then
For every $g: X \rightarrow A$ with $fg = 0$ we have a unique $\theta: X \rightarrow K$ with $i\theta = g$.
What is $g$? I have assumed $g$ is a morphism in the category $\mathcal{C}$ and thus I cannot use $i$ as the kernel of $f$ in $\mathcal{A}$ since I only have the properties for some morphisms $g$.