[proofplan]
We compare the defining presentation of $K_0(\mathcal A)$ with the defining universal property of the group completion of the direct-sum monoid. In the split exact structure, every conflation has middle object isomorphic to the direct sum of its kernel and cokernel, so every Grothendieck relation is a direct-sum relation. Conversely, every direct sum $A\oplus C$ gives a split conflation from $A$ to $A\oplus C$ to $C$, so all direct-sum relations are imposed in $K_0(\mathcal A)$. The universal property then identifies $K_0(\mathcal A)$ with the group completion of $M(\mathcal A)$.
[/proofplan]
[step:Construct the direct-sum monoid of isomorphism classes]
Let $M(\mathcal A)$ be the set of isomorphism classes of objects of $\mathcal A$. For an object $X$ of $\mathcal A$, write $\langle X\rangle\in M(\mathcal A)$ for its isomorphism class.
Because $\mathcal A$ is additive, it has a zero object $0$ and finite biproducts. Define an operation on $M(\mathcal A)$ by
\begin{align*}
\langle X\rangle+\langle Y\rangle=\langle X\oplus Y\rangle.
\end{align*}
This is well-defined because an isomorphism $X\cong X'$ and an isomorphism $Y\cong Y'$ induce an isomorphism $X\oplus Y\cong X'\oplus Y'$. The associativity, commutativity, and identity laws hold on isomorphism classes by the standard associativity, symmetry, and unit isomorphisms for biproducts. Hence $M(\mathcal A)$ is a commutative monoid with identity $\langle 0\rangle$.
[/step]
[step:Identify the Grothendieck relations with split direct-sum relations]
By definition of the Grothendieck group of an exact category, $K_0(\mathcal A)$ is generated by symbols $[X]$ for objects $X$ of $\mathcal A$, with the isomorphism relation $[X]=[Y]$ whenever $X\cong Y$, and with the exactness relation
\begin{align*}
[B]=[A]+[C]
\end{align*}
for every conflation
\begin{align*}
0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0
\end{align*}
in the chosen exact structure.
In the split exact structure, every conflation is split. Therefore, for each conflation as above, there is an isomorphism $B\cong A\oplus C$. Since $K_0(\mathcal A)$ identifies isomorphic objects, the conflation relation becomes
\begin{align*}
[A\oplus C]=[A]+[C].
\end{align*}
Thus every relation imposed by a split conflation is a direct-sum relation.
Conversely, for every pair of objects $A$ and $C$, the biproduct maps give a split conflation
\begin{align*}
0\longrightarrow A\longrightarrow A\oplus C\longrightarrow C\longrightarrow 0.
\end{align*}
The corresponding relation in $K_0(\mathcal A)$ is precisely
\begin{align*}
[A\oplus C]=[A]+[C].
\end{align*}
Therefore the relations defining $K_0(\mathcal A)$ are exactly the direct-sum relations on isomorphism classes.
[guided]
The point of this step is to check that the exact-category construction has not introduced any relation beyond the relation already visible in the direct-sum monoid.
By definition, $K_0(\mathcal A)$ is the abelian group generated by one symbol $[X]$ for each object $X$ of $\mathcal A$, subject first to the rule that isomorphic objects have the same class and second to the rule that every conflation
\begin{align*}
0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0
\end{align*}
contributes the relation
\begin{align*}
[B]=[A]+[C].
\end{align*}
Now use the special feature of the split exact structure. A conflation in this exact structure is split, so its middle object is isomorphic to the biproduct of the left and right terms. Thus, for a split conflation
\begin{align*}
0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0,
\end{align*}
there is an isomorphism $B\cong A\oplus C$. Since $K_0(\mathcal A)$ already identifies isomorphic objects, the relation
\begin{align*}
[B]=[A]+[C]
\end{align*}
is the same relation as
\begin{align*}
[A\oplus C]=[A]+[C].
\end{align*}
This proves that every exactness relation in $K_0(\mathcal A)$ is a direct-sum relation. We also need the reverse implication: every direct-sum relation must actually occur as a Grothendieck relation. Given objects $A$ and $C$, the biproduct $A\oplus C$ comes with the canonical inclusion $A\to A\oplus C$ and projection $A\oplus C\to C$. In the split exact structure these maps form a split conflation
\begin{align*}
0\longrightarrow A\longrightarrow A\oplus C\longrightarrow C\longrightarrow 0.
\end{align*}
The Grothendieck relation attached to this conflation is
\begin{align*}
[A\oplus C]=[A]+[C].
\end{align*}
Hence the two families of relations coincide: split conflations impose exactly the relations saying that the class of a direct sum is the sum of the classes.
[/guided]
[/step]
[step:Verify the universal property of the group completion]
Define
\begin{align*}
\iota:M(\mathcal A)&\longrightarrow K_0(\mathcal A)
\end{align*}
\begin{align*}
\langle X\rangle&\longmapsto [X].
\end{align*}
This map is well-defined because isomorphic objects have the same class in $K_0(\mathcal A)$. It is a monoid homomorphism because the direct-sum relation gives
\begin{align*}
\iota(\langle X\rangle+\langle Y\rangle)=\iota(\langle X\oplus Y\rangle)=[X\oplus Y]=[X]+[Y]=\iota(\langle X\rangle)+\iota(\langle Y\rangle),
\end{align*}
and the relation coming from the split conflation
\begin{align*}
0\longrightarrow 0\longrightarrow 0\longrightarrow 0\longrightarrow 0
\end{align*}
gives
\begin{align*}
[0]=[0]+[0].
\end{align*}
Since $K_0(\mathcal A)$ is an abelian group, cancellation gives $[0]=0$.
Let $H$ be an abelian group, and let
\begin{align*}
f:M(\mathcal A)\longrightarrow H
\end{align*}
be a monoid homomorphism. Define a function on the generators of $K_0(\mathcal A)$ by sending $[X]$ to $f(\langle X\rangle)$. This assignment respects isomorphism relations by the definition of $M(\mathcal A)$. It also respects every split conflation relation, because if
\begin{align*}
0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0
\end{align*}
is a split conflation, then $B\cong A\oplus C$, and hence
\begin{align*}
f(\langle B\rangle)=f(\langle A\oplus C\rangle)=f(\langle A\rangle+\langle C\rangle)=f(\langle A\rangle)+f(\langle C\rangle).
\end{align*}
Therefore the assignment descends to a unique [group homomorphism](/page/Group%20Homomorphism)
\begin{align*}
\overline f:K_0(\mathcal A)\longrightarrow H
\end{align*}
such that $\overline f([X])=f(\langle X\rangle)$ for every object $X$.
This homomorphism satisfies $\overline f\circ\iota=f$. It is unique with this property because the classes $[X]$ generate $K_0(\mathcal A)$ as an abelian group. Thus $\iota$ has the universal property of the group completion of $M(\mathcal A)$.
[/step]
[step:Conclude the identification]
The canonical monoid homomorphism
\begin{align*}
\iota:M(\mathcal A)\longrightarrow K_0(\mathcal A)
\end{align*}
has the universal property of the group completion. Therefore $K_0(\mathcal A)$ is canonically isomorphic to the group completion of the commutative monoid of isomorphism classes of objects of $\mathcal A$ under direct sum. This is the asserted identification.
[/step]
