If $\mathfrak{R}$ is an o-minimal expansion of an ordered field, then $K_0(\mathfrak{R})$ is a quotient of $\mathbb{Z}$. In a later post, we will see that, in fact, $K_0(\mathfrak{R})$ is exactly $\mathbb{Z}$.
Throughout this post $\mathfrak{R}$ is an o-minimal expansion of an ordered field.
Lemma 1: If $\mathfrak{R}$ is an expansion of an ordered field, then any two open intervals in $R$ are definably homeomorphic.
Proof: If $a < b$ are two elements of $R$, then the map $t \mapsto (1-t)a + tb$ defines a definable homeomorphism from $(0,1)$ to $(a,b)$. That is, all finite intervals are definably homeomorphic via linear maps.
For any $a \in R$, the maps $t \mapsto t + a$ defines a definable homeomorphsm between $(0,\infty)$ and $(a,\infty)$. Moreover, the map $t \mapsto -t$ defines a definable homeomorphism between $(-\infty,a)$ and $(-a,\infty)$. Thus, all half infinite intervals are definably homeomorphic. The map $t \mapsto \frac{t}{1-t}$ defines a definable homeomorphism between $(0,1)$ and $(0,\infty)$. Thus, all finite intervals and half infinite intervals are definably homeomorphic. Finally, the map given by $$t \mapsto \begin{cases} \frac{t}{1-t} & \text{ if } 1 > t > 0 \\ 0 & \text{ if } t= 0 \\ \frac{t}{1+t} & \text{ if } -1 < t < 0 \end{cases}$$ defines a definable homeomorphism between $(-1,1)$ and $R$ itself. Hence, all intervals are definably homeomorphic. $\Box$
Lemma 2: In $[(0,1)] = -1$ in $K_0(\mathfrak{R})$.
Proof: By Lemma 1, we know that $[(0,1)] = [(0,\frac{1}{2})] = [(\frac{1}{2},1)]$ in $K_0(\mathfrak{R})$ and in general for any singleton we have $[\{ \ast \}] = 1$ in the Grothendieck ring. Hence, $$ \begin{align} [(0,1)] &= [(0,\frac{1}{2}) \dot{\cup} \{ \frac{1}{2} \} \dot{\cup} (\frac{1}{2},1)] \\ &= [(0,\frac{1}{2})] + [\{ \frac{1}{2} \}] + [(\frac{1}{2},1)] \\ &= [(0,1)] + 1 + [(0,1)] \\ &= 2 [(0,1)] + 1 \end{align}$$ so that subtracting $[(0,1)] + 1$ from both sides we obtain $[(0,1)] = -1$. $\Box$
Proposition 3: If $C \subseteq R^n$ is a cell, then $C$ is definably homemorphic to $(0,1)^{\dim C}$.
Proof: We work by induction on $n$. When $n = 0$, then $C = \{ \ast \} = R^0$ which is also $(0,1)^0 = (0,1)^{\dim C}$. For the inductive case of $n+1$ either $C = \Gamma_{C’}(f)$ or $C = (f,g)_{C’}$ where $C’ \subseteq R^n$ is a cell and $f$ and $g$ are definable continuous functions on $C’$ (or possibly identically equal to $\pm \infty$ in the second case) with $f < g$. In the case that $C = \Gamma_{C’}(f)$, if $\pi:R^{n+1} \to R^n$ is the projection to the first $n$ coordinates, then $\pi \upharpoonright C:C \to C’$ is a definable homeomorphism and $\dim C = \dim C’$. Hence, by induction, $(0,1)^{\dim C} = (0,1)^{\dim C’} \approx C’ \approx C$ where we are writing
“$X \approx Y$” to mean that $X$ is definably homeomorphic to $Y$.
In the case that $C = (f,g)_{C’}$ we carry out the construction of Lemma 1 uniformly. Let $\rho:(0,1)^{\dim C’} \to C’$ be a definable homeomorphism. When both $f$ and $g$ take values in $R$, we define $\xi:(0,1)^{\dim C} \to C$ by $\langle x_1, \ldots, x_{\dim C’}, x_{\dim C} \rangle \mapsto \langle \rho(x_1, \ldots, x_{\dim C’}), (1 – x_{\dim C}) f(x_1, \ldots, x_{\dim C’}) + x_{\dim C} g(x_1, \ldots, x_{\dim C’})$. The other cases are similar and are left to the reader. $\Box$
Corollary 4: If $C$ is a cell, then in $K_0(\mathfrak{R})$ we have $[C] = (-1)^{\dim C}$.
Proof: By Lemma 3, we have $[C] = [(0,1)^{\dim C}]$ in $SK_0(\mathfrak{R})$. The multiplication in $K_0(\mathfrak{R})$ is induced by Cartesian product and by Lemma 2, we have $[(0,1)] = -1$ in $K_0(\mathfrak{R})$. Hence, $[C] = [(0,1)^{\dim C}] = [(0,1)]^{\dim C} = (-1)^{\dim C}$. $\Box$
Theorem 5: The unique map $\iota:\mathbb{Z} \to K_0(\mathfrak{R})$ is surjective.
Proof: We know that every element of $K_0(\mathfrak{R})$ is generated as a group by the classes of definable sets. Let $X \subseteq R^n$ be a definable set and let $\mathcal{C}$ be a cell decomposition of $R^n$ subjacent to $X$. Let $\mathcal{C}’ := \{C \in \mathcal{C} : C \subseteq X \}$. We then have $$\begin{align} [X] &= [\bigcup_{C \in \mathcal{C}’} X] \\ &= \sum_{C \in \mathcal{C}’} [C] \\ &= \sum_{C \in \mathcal{C}’} (-1)^{\dim C} \\ &\in \iota(\mathbb{Z}) \end{align}$$
Thus, as every element of $K_0(\mathfrak{R})$ is the difference of two effective classes, they all belong to the image of $\iota$. $\Box$