[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.[/step]

[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: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]