# Grothendieck group “commutes” with direct sum

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

• About the quotes in the title: I don't think there's any sketchiness is saying that ${\cal G}$ commutes with direct sum (or commutes up to an isomorphism in the appropriate category), but you can say that ${\cal G}$ distributes over direct sum if the other seems imprecise. – anomaly Jun 2 at 23:53
• The usual terminology is that $\mathcal{G}$ “respects” the direct sum, but “commutes” is almost accurate (technically, the direct sums on either side of the equal sign are different direct sums: one is the direct sum of commutative monoids, the other is the direct sum of abelian groups). – Arturo Magidin Jun 3 at 2:52

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.
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.