Group completions and the induced homomorphisms

Question

The group completion (aka Grothendieck group) of an abelian monoid $M$ is an abelian group $G(M)$ with a homomorphism $\iota:M \to G(M)$ of monoids satisfying the following universal property: for every homomorphism $f:M \to H$ whose target is an abelian group there exists a unique homomorphism $\phi:G(M)\to H$ of groups such that $\phi\circ\iota=f$.

By the universal property of $G$, every homomorphism $f:M \to N$ of monoids uniquely gives rise to a group homomorphism $G(f):G(M) \to G(N)$.

Of course, this concept is used to define $K$-theory: if $\mathcal V_X$ is the abelian monoid of the isomorphism classes of real vector bundle over a topological space $X$, then $KO(X):=G(\mathcal V_X)$.

In general, the homomorphism $\iota:M \to G(M)$ is not injective. (e.g. $M=\mathcal V_{\mathbb{CP}(1)}$.)

Questions:

1. Is there any criterion which implies (or is equivalent to) the injectivity of the map $\iota: M\to G(M)$? How about the cases when $M=\mathcal V_X$?
2. Is there any criterion for a homomorphism $f:M \to N$ of monoid which implies the injectivity of the group homomorphism $G(f):G(M) \to G(N)$? How about the cases when $f:=F^*:\mathcal V_X \to \mathcal V_Y$? (Here $F:Y \to X$ is a continuous map.)

Answer 1

Injectivity of the map $M \to G(M)$ is equivalent to cancellability in the monoid $M$: that $x + y = x + z$ implies $y = z$. Indeed, $G(M)$ may be constructed as the set of equivalence classes of pairs $(u, v)$ ("think $u - v$") where $(u, v) \sim (w, x)$ iff $\exists_t \; t + u + x = t + w + v$ in $M$; the existential condition is needed to establish transitivity. The universal map $i: M \to G(M)$ takes $u \in M$ to the class of $(u, 0)$. Given cancellabity in $M$ and $(u, 0) \sim (v, 0)$, we have $t + u = t + v$ for some $t$, whence $u = v$ by cancelling $t$; this shows injectivity. Conversely, if $M \to G(M)$ is injective and $t + u = t + v$ in $M$ for some $t$, then

$$i(t) + i(u) = i(t + u) = i(t + v) = i(t) + i(v)$$

in $G(M)$, whence $i(u) = i(v)$ since the group $G(M)$ is cancellable, whence $u = v$ in $M$ since $i$ is injective. This shows $M$ is cancellable.

I don't know of a very easy necessary and sufficient condition on $f: M \to N$ for $G(f)$ to be injective. Given that $N$ be cancellable, it is necessary and sufficient that $f$ be injective. (Injectivity of $f$ alone is not sufficient: consider the inclusion map of multiplicative monoids $(0, 1] \to [0, 1]$, where $G([0, 1])$ is the trivial group since $0$ is absorbing in $[0, 1]$. Similarly, the homomorphism $[0, 1] \to \{1\}$ shows that injectivity of $f$ isn't necessary either.) A slightly more sophisticated sufficient condition is that $f$ be injective and that $\forall_{t \in N} \exists_{s \in N, m \in M} \; s + t = f(m)$. But it's hard to say how useful that would be.

I'd have to think a little on the topological $K$-theory sub-question.