[proofplan]
We use the presentation of $K_0(\mathcal E)$ by generators and relations. The generators are isomorphism classes of objects, and the relations are exactly the conflation relations $[B]-[A]-[C]$. The displayed additivity relation follows immediately from this quotient presentation. For the universal property, an isomorphism-invariant additive function extends to the free abelian group on isomorphism classes, kills all defining relations, and therefore descends uniquely to the quotient.
[/proofplan]
[step:Present $K_0(\mathcal E)$ by generators and conflation relations]
Let $\operatorname{Iso}(\mathcal E)$ denote the set of isomorphism classes of objects of $\mathcal E$. For an object $X$ of $\mathcal E$, let $\langle X\rangle\in \operatorname{Iso}(\mathcal E)$ denote its isomorphism class.
Let $F$ be the free abelian group on $\operatorname{Iso}(\mathcal E)$. For each $\alpha\in\operatorname{Iso}(\mathcal E)$, write $e_\alpha\in F$ for the corresponding basis element. Thus every element of $F$ is a finite integer linear combination of the elements $e_\alpha$.
Let $R\leq F$ be the subgroup generated by all elements
\begin{align*}
e_{\langle B\rangle}-e_{\langle A\rangle}-e_{\langle C\rangle}
\end{align*}
where
\begin{align*}
0 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 0
\end{align*}
is a conflation in $\mathcal E$. By the presentation-definition of the Grothendieck group of an exact category,
\begin{align*}
K_0(\mathcal E)=F/R.
\end{align*}
Let
\begin{align*}
q:F\to F/R
\end{align*}
be the quotient homomorphism. For an object $X$ of $\mathcal E$, the class $[X]\in K_0(\mathcal E)$ is, by definition,
\begin{align*}
[X]=q(e_{\langle X\rangle}).
\end{align*}
[/step]
[step:Read the additivity relation in the quotient]
Let
\begin{align*}
0 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 0
\end{align*}
be a conflation in $\mathcal E$. By the definition of $R$, the element
\begin{align*}
e_{\langle B\rangle}-e_{\langle A\rangle}-e_{\langle C\rangle}
\end{align*}
belongs to $R$. Applying the quotient homomorphism $q:F\to F/R$, we obtain
\begin{align*}
q(e_{\langle B\rangle})-q(e_{\langle A\rangle})-q(e_{\langle C\rangle})=0.
\end{align*}
Using the definition $[X]=q(e_{\langle X\rangle})$, this is
\begin{align*}
[B]-[A]-[C]=0.
\end{align*}
Equivalently,
\begin{align*}
[B]=[A]+[C].
\end{align*}
[/step]
[step:Extend an additive invariant to the free abelian group]
Let $G$ be an abelian group, and let
\begin{align*}
f:\operatorname{Ob}(\mathcal E)\to G
\end{align*}
be invariant under isomorphism and additive on conflations. Since $f$ is invariant under isomorphism, there is a well-defined function
\begin{align*}
f_{\operatorname{iso}}:\operatorname{Iso}(\mathcal E)\to G
\end{align*}
given by
\begin{align*}
f_{\operatorname{iso}}(\langle X\rangle)=f(X).
\end{align*}
The value does not depend on the chosen representative $X$ because isomorphic objects have the same $f$-value.
By the universal property of the free abelian group $F$, there is a unique [group homomorphism](/page/Group%20Homomorphism)
\begin{align*}
\widetilde f:F\to G
\end{align*}
such that, for every object $X$ of $\mathcal E$,
\begin{align*}
\widetilde f(e_{\langle X\rangle})=f(X).
\end{align*}
[/step]
[step:Show the extended homomorphism kills every defining relation]
We prove that $R\subseteq \ker(\widetilde f)$. Since $R$ is generated by the conflation relations, it is enough to check the value of $\widetilde f$ on each generator of $R$.
Let
\begin{align*}
0 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 0
\end{align*}
be a conflation in $\mathcal E$. Then
\begin{align*}
\widetilde f(e_{\langle B\rangle}-e_{\langle A\rangle}-e_{\langle C\rangle})
=
f(B)-f(A)-f(C).
\end{align*}
By the assumed additivity of $f$ on conflations,
\begin{align*}
f(B)=f(A)+f(C).
\end{align*}
Therefore
\begin{align*}
\widetilde f(e_{\langle B\rangle}-e_{\langle A\rangle}-e_{\langle C\rangle})=0.
\end{align*}
Thus every generator of $R$ lies in $\ker(\widetilde f)$, and hence $R\subseteq \ker(\widetilde f)$.
[guided]
The point of this step is to check exactly why the map from the free group can pass to the quotient. A homomorphism out of a quotient $F/R$ is the same thing as a homomorphism out of $F$ that sends every element of $R$ to $0$. Thus we must verify that the subgroup $R$ of defining relations is contained in the kernel of $\widetilde f$.
By definition, $R$ is generated by all elements
\begin{align*}
e_{\langle B\rangle}-e_{\langle A\rangle}-e_{\langle C\rangle}
\end{align*}
coming from conflations
\begin{align*}
0 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 0.
\end{align*}
It is therefore enough to evaluate $\widetilde f$ on one such generator. Since $\widetilde f:F\to G$ is a group homomorphism and satisfies $\widetilde f(e_{\langle X\rangle})=f(X)$ for every object $X$, we get
\begin{align*}
\widetilde f(e_{\langle B\rangle}-e_{\langle A\rangle}-e_{\langle C\rangle})
=
f(B)-f(A)-f(C).
\end{align*}
The hypothesis that $f$ is additive on conflations says precisely that
\begin{align*}
f(B)=f(A)+f(C).
\end{align*}
Substituting this equality gives
\begin{align*}
f(B)-f(A)-f(C)=0.
\end{align*}
Hence every defining conflation relation maps to $0$ under $\widetilde f$. Since kernels are subgroups and $R$ is generated by these relations, the whole subgroup $R$ is contained in $\ker(\widetilde f)$.
[/guided]
[/step]
[step:Descend to $K_0(\mathcal E)$ and prove uniqueness]
Since $R\subseteq \ker(\widetilde f)$, the [universal property of quotient groups](/theorems/2701) gives a unique group homomorphism
\begin{align*}
\overline f:F/R\to G
\end{align*}
such that
\begin{align*}
\overline f\circ q=\widetilde f.
\end{align*}
Using $K_0(\mathcal E)=F/R$, we regard this as a homomorphism
\begin{align*}
\overline f:K_0(\mathcal E)\to G.
\end{align*}
For every object $X$ of $\mathcal E$,
\begin{align*}
\overline f([X])
=
\overline f(q(e_{\langle X\rangle}))
=
\widetilde f(e_{\langle X\rangle})
=
f(X).
\end{align*}
It remains to prove uniqueness. Let
\begin{align*}
h:K_0(\mathcal E)\to G
\end{align*}
be a group homomorphism satisfying $h([X])=f(X)$ for every object $X$ of $\mathcal E$. The elements $[X]$, as $X$ ranges over the objects of $\mathcal E$, generate $K_0(\mathcal E)$ because $K_0(\mathcal E)$ is the quotient of the free abelian group generated by the corresponding isomorphism classes. Therefore $h$ and $\overline f$ agree on a generating set of $K_0(\mathcal E)$, so $h=\overline f$. This proves both the required factorization and its uniqueness.
[/step]
