I want to assess the following claim:
If $F \in k[x_{0},...,x_{n}]$ is homogeneous and $F=x_{0}^{\alpha}G$ (where $x_{0}$ does not divide $G$) then $\beta \circ \alpha(F)=G$.
Here $\alpha$ denotes the dehomogeneization map with respect to $x_0$: $\alpha(f) = f(1, x_1, \ldots, x_n),$while $\beta$ is the homogeneization map for the same variable: $\beta(f) = x_0^{d(G)} f(x_0/x_0, \ldots, x_n/x_0).$Here $d(f)$ denotes the degree of the polynomial $f$.
Attempt: I tried with the following:
$\beta \circ \alpha(F)(x_{0},x_{1},...,x_{n})=x_{0}^{d(G)}G(1,\frac{x_{1}}{x_{0}},\frac{x_{2}}{x_{0}},...,\frac{x_{n}}{x_{0}}).$
So why do we have the equality $x_{0}^{d(G)}G(1,\frac{x_{1}}{x_{0}},\frac{x_{2}}{x_{0}},...,\frac{x_{n}}{x_{0}}) = G$?