# Show that $K_2(F)$ is a direct summand of $K_2F(t)$

I have a question regarding Example 6.1.2 (page 252) from the book "The K-Book" by Charles Weibel. Here is the statement:

Example 6.1.2:

Let $$F(t)$$ be a rational function field in one variable $$t$$ over $$F$$. Then $$K_2(F)$$ is a direct summand of $$K_2F(t)$$.

$$\text{ }$$

I find it quite difficult to understand the argument, so I hope I can get some help from here.

His goal is to construct an inverse map $$\lambda: K_2 F(t)\to K_2(F)$$ to the natural map $$K_2(F)\to K_2 F(t)$$.

Make the following definitions $$f(t)=\frac{a_0t^n+\cdots + a_n}{b_0t^m+\cdots+ b_m},$$ $$\operatorname{lead}(f):=\frac{a_0}{b_0}$$ and $$\lambda(\{f,g\})=\{\operatorname{lead}(f),\operatorname{lead}(g)\}$$.

Now he says that we may check the representation of Matsumoto's Theorem, to show that $$\lambda$$ is a homomorphism.

We begin to show bilinearity. Let $$f,f',g\in F(t)^\times$$ then $$\lambda(\{f,g\}\{f',g\})=\lambda(\{ff',g\})=\{\operatorname{lead}(ff'),g\}=\{\operatorname{lead}(f)\operatorname{lead}(f'),g\}=\{\operatorname{lead}(f),g\}\{\operatorname{lead}(f'),g\}=\lambda(\{f,g\})\lambda(\{f',g\}).$$

By symmetry, the same argument works if we consider something like $$\lambda(\{f,g\}\{f,g'\})$$ instead.

Let us now see that the Steinberg identity is, also, satisfied. Consider $$f\in F(t)^\times$$ and assume $$m, then we have that $$\operatorname{lead}(1-f)=-\operatorname{lead}(f)$$, since $$1-\frac{a_0t^n+\cdots a_{n-m}t^{m}+\cdots + a_n}{b_0t^m+\cdots + b_m}=\frac{b_0t^m+\cdots + b_m}{b_0t^m+\cdots + b_m}-\frac{a_0t^n+\cdots a_{n-m}t^{m}+\cdots + a_n}{b_0t^m+\cdots + b_m}=\frac{-(a_0t^n+\cdots a_{n-(m+1)}t^{m+1})+(b_m-a_{n-m})t^{m}+\cdots + (b_m-a_n)}{b_0t^m+\cdots + b_m},$$ and so $$\operatorname{lead}(1-f)=-\operatorname{lead}(f)$$. It was showed earlier in the book that $$\{x,-x\}=1$$ for every $$x\in E^\times$$, where $$E$$ is some field. Thus $$\{\operatorname{lead}(f),\operatorname{lead}(1-f)\}=\{\operatorname{lead}(f),-\operatorname{lead}(f)\}=1$$. One can also show by similar reasoning that $$\{\operatorname{lead}(f),\operatorname{lead}(1-f)\}=\{\operatorname{lead}(f),1-\operatorname{lead}(f)\}=1$$ if $$n=m$$ and that $$\{\operatorname{lead}(f),\operatorname{lead}(1-f)\}=\{\operatorname{lead}(f),1\}$$ if $$n.

Lastly, they say that since $$K_2$$ commutes with filtered colimits, it follows that $$K_2(F)$$ injects into $$K_2F(T)$$ for every purely transcendental extension $$F(T)$$ of $$F$$.

Questions:

1. Why does checking the criteria for Matsuomoto's Theorem prove that $$\lambda$$ is a homomorphism? Don't we want to pick symbols $$\{f,g\},\{f',g'\}$$ and check that it is a group homomorphism: $$\lambda(\{f,g\}\{f',g'\})=\lambda(\{f,g\})\lambda(\{f',g'\})?$$ By checking Matsumoto's Theorem don't we just make sure that we map elements from $$K_2F(t)$$ to $$K_2(F)$$?
2. Why does the following equality hold $$\{\operatorname{lead}(f),1\}=1$$?
3. How comes that the last paragraph implies that $$K_2(F)$$ injects into $$K_2F(T)$$? I don't really understand why $$K_2$$ commutes with filtered colimits but if I take it for granted for the moment, why does it prove the fact?
4. Lastly, why does showing that $$K_2(F)\to K_2 F(T)$$ is injective show that $$K_2(F)$$ is a direct summand of $$K_2F(t)$$?

The answers are maybe/probably easy, but I am quite confused right now, new to $$K$$-theory and I don't manage to figure it out.

Best wishes,

Joel

Edit:

Matsumoto's Theorem (for completeness):

If $$F$$ is a field, then $$K_2(F)$$ is the abelian group generated by the set of Steinberg symbols $$\{x,y\}$$ with $$x,y\in F^\times$$, subject to (only) the following relations:

$$\quad\quad\text{(Bilinearity)}\quad\{xx',y\}=\{x,y\}\{x',y\}\text{ and }\{x,yy'\}=\{x,y\}\{x,y'\}$$ $$\text{(Steinberg identity)}\quad\{x,1-x\}=1\text{ for all }x\not = 0,1.$$