Grothendieck Group of Category of finite dim space

Let $k^f$$k^f$ be the category of finite-dimensional vector spaces over $k$$k$, and $\sim$$\sim$ be the isomorphism relationship.

We use $[V]$$[V]$ to denote the isomorphic class of vector space, i.e. $[V]=[U]⟺V\cong U$$[V]=[U]\iff V\cong U$.

Consider the free $\mathbb{Z}$$\mathbb Z$ module $F(k^f)$$F(k^f)$ generated by those $[V]$$[V]$ and the submodule $E$$E$ generated by $[V]-[U]-[W]$$[V]-[U]-[W]$, where

$$\begin{array}{}\text{(1)}& 0\stackrel{}{\to }U\stackrel{}{\to }V\stackrel{}{\to }W\stackrel{}{\to }0\end{array}$$

is an exact sequence.

Then we define the Grothendieck Group $K_0(k^f)$$K_0(k^f)$ to be ${\displaystyle \frac{F(k^f)}{E}}$$\dfrac{F(k^f)}{E}$. Hence we have $[U\oplus W]=[U]+[W]$$[U\oplus W]=[U]+[W]$ in $K_0$$K_0$​.

Lemma. Every element in $K_0(k^f)$$K_0(k^f)$ could be written as $[A]-[B]$$[A]-[B]$.

Proof. If $s\in K_0(k^f),$$s\in K_0(k^f),$ then $s={\lambda }_1[V_1]+{\lambda }_2[V_2]+...+{\lambda }_n[V_n]$$s=\lambda_1[V_1]+\lambda_2[V_2]+...+\lambda_n[V_n]$. WLOG, assume that ${\lambda }_1$$\lambda_1$ to ${\lambda }_i$$\lambda_i$ are positive and ${\lambda }_{i+1}$$\lambda_{i+1}$ to ${\lambda }_n$$\lambda_n$ are negative, then $s=[V_1^{{\lambda }_1}\oplus V_2^{{\lambda }_2}\oplus ...V_i^{{\lambda }_i}]-[V_{i+1}^{{\lambda }_{i+1}}\oplus V_{i+2}^{{\lambda }_{i+2}}\oplus ...V_n^{{\lambda }_n}]$$s=[V_1^{\lambda_1}\oplus V_2^{\lambda_2}\oplus...V_i^{\lambda_i}]-[V_{i+1}^{\lambda_{i+1}}\oplus V_{i+2}^{\lambda_{i+2}}\oplus...V_n^{\lambda_n}]$. $◻$$\square$

Let $A$$A$ be a $\mathbb{Z}$$\Z$-module and $\delta :K_0(k^f)\to A$$\delta:K_0(k^f)\to A$ be a function satisfies $\delta (V/U)=\delta (V)-\delta (U)$$\delta(V/U)=\delta(V)-\delta(U)$, i.e.,

$$\begin{array}{}\text{(2)}& \delta (V)-\delta (U)-\delta (V/U)=0\end{array}$$

Then it induce an universal group homomorphism $\delta :K_0(k^f)\to A$$\delta:K_0(k^f)\to A$.

Well, observe that $\dim :K_0(k^f)\to \mathbb{Z}$$\dim:K_{0}(k^f)\to\mathbb Z$ satisfies $\dim (V/U)=\dim (V)-\dim (U)$$\dim(V/U)=\dim(V)-\dim(U)$, we have the following proposition.

Proposition. $K_0(k^f)\cong \mathbb{Z}$$K_0(k^f)\cong\mathbb Z$, and the isomorphism is induced by $\dim .$$\dim.$

Proof. Easy to see $\dim :K_0\to \mathbb{Z}$$\dim:K_0\to\mathbb Z$ is surjective. To see it is injective, we prove that the kernel is $0$$0$.

By the previous lemma we know that every $s\in K_0$$s\in K_0$ can be written as $[A]-[B]$$[A]-[B]$.

Hence

$$\begin{array}{}\text{(3)}& \dim ([A]-[B])=0=\dim ([A])-\dim ([B])⟺\dim ([A])=\dim ([B])\end{array}$$

Also, by the property of $\dim$$\dim$ we have:

$$\begin{array}{}\text{(4)}& \dim ([A])=\dim ([B])⟺A\cong B⟺[A]=[B]⟺[A]-[B]=0◻\end{array}$$