**Proof.**
Let $x \in F(B)$. The map $p : B \to B[t, t^{-1}]$ is free hence we know that

\[ F(B[t, t^{-1}]) = \bigoplus \nolimits _{k \in \mathbf{Z}} F(p)(F(B)) \cdot t^ k = \bigoplus \nolimits _{k \in \mathbf{Z}} F(B) \cdot t^ k \]

as indicated we drop the $F(p)$ in the rest of the proof. Write $F(a)(x) = \sum t^ k x_ k$ for some $x_ k \in F(B)$. Denote $\epsilon : B[t, t^{-1}] \to B$ the $B$-algebra map $t \mapsto 1$. Note that the compositions $\epsilon \circ p, \epsilon \circ a : B \to B[t, t^{-1}] \to B$ are the identity. Hence we see that

\[ x = F(\epsilon )(F(a)(x)) = F(\epsilon )(\sum t^ k x_ k) = \sum x_ k. \]

On the other hand, we claim that $x_ k \in F(B)^{(k)}$. Namely, consider the commutative diagram

\[ \xymatrix{ B \ar[r]_ a \ar[d]_{a'} & B[t, t^{-1}] \ar[d]^ f \\ B[s, s^{-1}] \ar[r]^-g & B[t, s, t^{-1}, s^{-1}] } \]

where $a'(b) = s^{\deg (b)}b$, $f(b) = b$, $f(t) = st$ and $g(b) = t^{\deg (b)}b$ and $g(s) = s$. Then

\[ F(g)(F(a'))(x) = F(g)(\sum s^ k x_ k) = \sum s^ k F(a)(x_ k) \]

and going the other way we see

\[ F(f)(F(a))(x) = F(f)(\sum t^ k x_ k) = \sum (st)^ k x_ k. \]

Since $B \to B[s, t, s^{-1}, t^{-1}]$ is free we see that $F(B[t, s, t^{-1}, s^{-1}]) = \bigoplus _{k, l \in \mathbf{Z}} F(B) \cdot t^ ks^ l$ and comparing coefficients in the expressions above we find $F(a)(x_ k) = t^ k x_ k$ as desired.

Finally, the image of $F(B_0) \to F(B)$ is contained in $F(B)^{(0)}$ because $B_0 \to B \xrightarrow {a} B[t, t^{-1}]$ is equal to $B_0 \to B \xrightarrow {p} B[t, t^{-1}]$.
$\square$

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