Grothendieck group "commutes" with direct sum

Question

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 $\bigoplus_{\alpha} M_{\alpha}$ ?

Answer 1

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.

Answer 2

The Grothendieck completion is the left adjoint of the forgetful functor from Abelian groups to commutative monoids. That is, if $\mathcal{M}$ denotes the functor $\mathcal{M}\colon\mathfrak{A}\to\mathfrak{M}$ from abelian groups to commutative monoids that maps the abelian group $G$ to itself considered as a monoid, then for any commutative monoid $M$ and abelian group $G$, we have a natural isomorphism $$\mathfrak{M}(M,\mathcal{M}(A))\cong \mathfrak{A}(\mathcal{G}(M),A).$$

Left adjoints respect colimits, right adjoints respect limits. As the direct sum is a coproduct/colimit, it follows that $\mathcal{G}$ respects direct sums. Explicitly, recall that a map from a direct sum is equivalent to maps from each constituent: each morphism $f\colon\oplus_{\alpha\in A}X_{\alpha}\to Z$ corresponds to a family of maps $\{ f_{\alpha}\colon X_{\alpha}\to Z\}_{\alpha\in A}$ (in any category in which the direct sum is a coproduct; if not, then you should use the coproduct instead of the direct sum). Thus, for every abelian group $A$, $$\begin{align*} \mathfrak{A}(\mathcal{G}(\oplus_{\alpha}M_{\alpha}),A) &\cong \mathfrak{M}(\oplus_{\alpha}M_{\alpha},\mathcal{M}(A))\\ &\cong \prod_{\alpha}\mathfrak{M}(M_{\alpha},\mathcal{M}(A))\\ &\cong \prod_{\alpha}\mathfrak{A}(\mathcal{G}(M_{\alpha}),A)\\ &\cong \mathfrak{A}(\oplus_{\alpha}\mathcal{G}(M_{\alpha}),A). \end{align*}$$ This means that $\mathcal{G}(\oplus_{\alpha} M_{\alpha})$ has the universal property of $\oplus_{\alpha}\mathcal{G}(M_{\alpha})$, hence the two are isomorphic.

Cf. the proof that the free monoid on a disjoint union of two sets is the coproduct of the free monoids on each set. It’s the same, because the relation between the “free monoid” construction and the “underlying set” functor is the same as the relation between the “Grothendieck completion group” construction and the “underlying monoid” functor. This in turn is a special case of the result mentioned above, that left adjoints respect colimits and right adjoints respect limits. Which is one reason why you should care about adjoint functors.