Graded algebras and symbols

Solution 1:

$\operatorname{Symb}$ is only injective or surjective or an additive map in degenerate cases (like $A_0=A$). In case $A$ is graded and the filtration on $A$ is the natural one described by the OP, $\operatorname{Symb}$ is only injective or surjective or an additive map if the grading is concentrated in degree $0$. In the graded case, $A$ is isomorphic (even graded isomorphic) to $\operatorname{Gr} A$, but not via the map $\operatorname{Symb}$.

As an example, suppose $A=F[x]$ for a ring $F$, with the usual grading (elements of $F$ have degree $0$, $\deg x=1$). Then $\operatorname{Symb}(a)$ is the leading term of $a$. (For example, $\operatorname{Symb}(3x^2+2x-5)=3x^2$.) We see $x^2\ne x^2-1$, but $\operatorname{Symb}(x^2)=\operatorname{Symb}(x^2-1)$, and we see $\operatorname{Symb}(a)=x^2+1$ is impossible. Thus injectivity and surjectivity both fail. There are two sorts of counterexamples to additivity, namely (i) $\operatorname{Symb}((x^2)+(1-x^2))=1$, while $\operatorname{Symb}(x^2)+\operatorname{Symb}(1-x^2)=0$ and (ii) $\operatorname{Symb}(x^2+x)=x^2$, while $\operatorname{Symb}(x^2)+\operatorname{Symb}(x)=x^2+x$.