Weibel Lemma 1.6.2. The following is (a part of) Lemma 1.6.2. from Weibel's Homological Algebra.$\newcommand{\C}[1]{\mathcal{#1}}\DeclareMathOperator{\coker}{coker}\newcommand{\md}[1]{{\left\lvert #1 \right\lvert}}\DeclareMathOperator{\Hom}{Hom}$

Let $\C{C} \subset \C{A}$ be a full subcategory of an abelian category $\C{A}$.
$\C{C}$ is additive $\Leftrightarrow$ $0 \in \C{C}$, and $\C{C}$ is closed under $\oplus$.


I am having difficulty with the $\Rightarrow$ direction.
First, here is what I interpret the right side to mean: $\C{C}$ contains a zero object of $\C{A}$. Furthermore, given two objects $A, B \in \C{C}$, $\C{C}$ contains a coproduct of $A$ and $B$. (Where this coproduct was their coproduct in $\C{A}$.)
The part about the zero object was easy to show. Indeed, since $\C{C}$ is additive, it has some zero object, say $0'$. We now have to show that $0'$ is also a zero object in $\C{A}$. This follows since $\C{C}$ is a full subcategory because we have
\begin{equation*} 
\md{\Hom_{\C{A}}(0', 0')} = \md{\Hom_{\C{C}}(0', 0')} = 1,
\end{equation*}
showing that $0'$ is a zero object in $\C{A}$ as well.
The same sort of trick does not seem to work for $\oplus$, though. I started with objects $A, B \in \C{C}$. Now, since $\C{C}$ is additive, the coproduct $A \oplus' B$ in $\C{C}$ exists. I believe that I need to show that this is in $\C{A}$ as well but I don't see how.
I believe that I really need to use abelian-ity of $\C{A}$ since otherwise this statement is not true. That is, just being full doesn't mean that a coproduct in the subcategory is also one in the larger category. (Consider the full subcategory of abelian groups within groups. The coproduct in the former isn't one in the latter.)
 A: The non-trivial statement you need is that if $\mathcal{A}$ is a category admitting finite biproducts and $\mathcal{C}\rightarrow\mathcal{A}$ is the inclusion of a subcategory admitting finite biproducts, then that inclusion preserves the finite biproducts. To prove this, take a collection of objects $X_1,\dots,X_n\in\mathcal{C}$ and denote the coproducts $X_1\oplus_{\mathcal{C}}\dots\oplus_{\mathcal{C}}X_n$ and $X_1\oplus_{\mathcal{A}}\dots\oplus_{\mathcal{A}}X_n$ respectively. These come with natural inclusion and projection morphisms, which, by the universal properties of $X_1\oplus_{\mathcal{A}}\dots\oplus_{\mathcal{A}}X_n$ as both product and coproduct, induce natural morphisms $X_1\oplus_{\mathcal{C}}\dots\oplus_{\mathcal{C}}X_n\rightarrow X_1\oplus_{\mathcal{A}}\dots\oplus_{\mathcal{A}}X_n$ and $X_1\oplus_{\mathcal{A}}\dots\oplus_{\mathcal{A}}X_n\rightarrow X_1\oplus_{\mathcal{C}}\dots\oplus_{\mathcal{C}}X_n$. Using, again, the universal property and calculating the components of their composites using the respective inclusions/projections (here, one needs the compatibility of inclusions and projections in a biproduct), one confirms that these maps are in fact inverse. Since the calculation in particular demonstrates that the canonical projections correspond under these isomorphisms, this shows that $X_1\oplus_{\mathcal{C}}\dots\oplus_{\mathcal{C}}X_n$ is also a biproduct in $\mathcal{A}$, as desired. Note that this argument works in particular if $n=0$, which is the case of the zero object.
