The Grothendieck completion group of a commutative monoid $M$ is the unique (up to isomorphism) pair $\langle \mathcal{G}(M), i_M\rangle$, where $\mathcal{G}(M)$ is an abelian group and $i_M\colon M\to\mathcal{G}(M)$ is a monoid homomorphism, satisfying the universal property: for every abelian group $G$ and monoid homomorphism $f\colon M\to G$ there exists a unique $\varphi\colon\mathcal{G}(M)\to G$ such that $f = \varphi\circ i_M$.
Let $M$ and $N$ be commutative monoids. It's easily seen that $M\oplus N$ is a commutative monoid with component-wise operation.
Question: Is it true that $\mathcal{G}(M\oplus N) \cong \mathcal{G}(M)\oplus\mathcal{G}(N)$ ?
The universal property applied to the monoid homomorphism $i_{M}\oplus i_{N}\colon M\oplus N\to\mathcal{G}(M)\oplus\mathcal{G}(N)$ gives a group homomorphism $\varphi\colon\mathcal{G}(M\oplus N)\to\mathcal{G}(M)\oplus\mathcal{G}(N)$ such that $i_{M}\oplus i_{N} = \varphi\circ i_{M\oplus N}$ and I was trying to prove that $\varphi$ is the desired isomorphism, without success.
Is the answer to the question affirmative? If so, is this the correct approach?
Any hints would be appreciated. Thanks in advance.
EDIT: Also, is it true if we replace $M\oplus N$ by $\bigoplus_{\alpha} M_{\alpha}$ ?
Best Answer
Here's a sketch. You have to construct the inverse using universal properties as well. You have a composition $$M \hookrightarrow M\oplus N \stackrel{i_{M\oplus N}}{\longrightarrow} \mathcal{G}(M\oplus N)$$which induces a map $\mathcal{G}(M) \to \mathcal{G}(M\oplus N)$. Similarly, you get a map $\mathcal{G}(N) \to \mathcal{G}(M\oplus N)$. The universal property of the direct sum joins these two maps into a map $\psi\colon\mathcal{G}(M)\oplus \mathcal{G}(N) \to \mathcal{G}(M\oplus N)$. Now you have two maps $$\psi\circ \varphi\colon \mathcal{G}(M\oplus N) \to \mathcal{G}(M\oplus N)\quad\mbox{and}\quad\varphi\circ \psi\colon \mathcal{G}(M)\oplus \mathcal{G}(N) \to \mathcal{G}(M)\oplus \mathcal{G}(N).$$Using the uniqueness provided by the universal properties of $\mathcal{G}$ and $\oplus$, argue that these compositions equal the identity. It works the same for defining maps $$\bigoplus_{\alpha} \mathcal{G}(M_\alpha) \to \mathcal{G}\left(\bigoplus_{\alpha}M_\alpha\right) \quad\mbox{and}\quad \mathcal{G}\left(\bigoplus_\alpha M_\alpha\right) \to \bigoplus_\alpha \mathcal{G}(M_\alpha)$$and running the above argument through.