Groupification and perfection of a commutative monoid
(1) If $M \to M^{pf}$ is not injective, there exist $x \neq y$ and $n\in \mathbb{N}_{\geq 1}$ such that $f_{n1}(x) = f_{n1}(y)$. (The fact that elements are equal in the colimit iff they are equal in some term requires some thought. It is not a general property; it follows here from the structure of the diagram defining the colimit.) So let us assume that $x,y\in M$ satisfy $nx = ny$.
Let $Z = [x]  [y]$ in $M^{gp}$. We have $nZ = [nx]  [ny] = 0$, which lies in the image $\psi(M)$, so saturation implies that $Z \in \psi(M)$. Say $Z=\psi(z)$ for $z\in M$. Since $Z$ satisfies $Z + [y] = [x]$, integrality implies that $z + y = x$. The element $z$ satisfies $\psi(nz)=nZ=0\in M^{gp}$, so integrality implies $nz = 0\in M$. Since $M$ is torsionfree this implies $z = 0$, so $z+y=x$ becomes $y = x$. This shows that $M\to M^{pf}$ is injective.
(2) Since $M^{pf}$ is perfect (an inverse to $x\mapsto nx$ is the map induced by sending $x\in M_a$ to $x\in M_{na}$), an isomorphism $M\to M^{pf}$ implies that $M$ is perfect. Conversely, if $M$ is perfect, we can change coordinates on $M_a$ by $x \mapsto \frac{1}{n}x$ (the inverse of $x\mapsto nx$, which is automatically a morphism). Now the direct system is constant and so $M\to M_1\to M^{pf}$ is an isomorphism.
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Makoto Kato
Updated on August 01, 2022Comments

Makoto Kato over 1 year
The following definitions and proposition are taken from the paper The geometry of Frobenioids I by S. Mochizuki.
Let $\mathbb{N}_{\ge 1}$ be the set of positive integers. $\mathbb{N}_{\ge 1}$ is a directed set by the order relation $ab$. Let $M$, $N$ be commutative monoids(the binary operations are written additively). A map $f\colon M \rightarrow N$ is called a morphism if $f(x + y) = f(x) + f(y)$ for any $x, y$ and $f(0) = 0$.
For every $n \in \mathbb{N}_{\ge 1}$, the multiplication map $x \rightarrow nx$ on $M$ is a morphism.
$M$ is said to be torsionfree if $nx = 0$ implies $x = 0$ for every $n \in \mathbb{N}_{\ge 1}$.
Let $M^{gp}$ be the groupification of $M$, i.e. the grothendieck group of $M$. Let $\psi\colon M \rightarrow M^{gp}$ be the canonical map. If $\psi$ is injective, $M$ is said to be integral.
$M$ is said to be saturated if $na \in \psi(M)$ for some $n \in \mathbb{N}_{\ge 1}$ and $a \in M^{gp}$, then $a \in \psi(M)$.
$M$ is said to be perfect if the multiplication map $x \rightarrow nx$ is bijective for every $n \in \mathbb{N}_{\ge 1}$.
The perfection $M^{pf}$ of $M$ is defined as follows. Let $M_a = M$ for every $a \in \mathbb{N}_{\ge 1}$. if $ab$, we define a morphism $f_{ba}\colon M_a \rightarrow M_b$ by $f_{ba}(x) = (b/a)x$. Then $(M_a)_{a\in \mathbb{N}_{\ge 1}}$ and $(f_{ba})$ consitute a direct(or inductive) system. We define $M^{pf} = colim_{a\in \mathbb{N}_{\ge 1}} M_a$. There exists the canonical morphism $\phi\colon M = M_1 \rightarrow M^{pf}$.
My question: How do we prove the following proposition?
Proposition
(1) $\phi\colon M \rightarrow M^{pf}$ is injective if $M$ is torsionfree, integral, and saturated.
(2) $M$ is perfect if and only if $\phi$ is an isomorhism.
Motivation Recently(August, 2012), S. Mochizuki submitted a series of papers(Interuniversal Teichmuller Theory I,II,III,IV) which develops his new theory. As an application of his theory, he wrote a proof of ABC conjecture. I think the validity of the proof has not yet been confirmed by other mathematicians. However, considering his track record, I think it's worthwhile to read the papers. He referred to The geometry of Frobenioids I for the notation and terminolgy concerning monoids and categories.

Makoto Kato almost 10 years@user18921 I'm afraid I don't understand what you mean. Could you rephrase it?


Berci about 11 yearsI got stucked, too.. sorry, Turn back on that later.

Makoto Kato about 11 yearsI think (1) is correct. However I don't understand (2) fully. Could you elaborate on it?

Tom Church about 11 yearsIf $M$ is perfect, then each map $f_{ba}\colon M_a\to M_b$ is an isomorphism of monoids. Therefore the colimit is boring: it is isomorphic to $M_1$. (To see this, check that for any $a\in \mathbb{N}_{\geq 1}$ and any $x\in M_a$, there exists some $y\in M_1$ so that $f_{a1}(y)=x$. This is worth working out.)