Take $A$ a commutative unitary $\mathbb{C}$-algebra and take $A_0\subset A_1\subset...$ a filtration on $A$. If $$GrA=\bigoplus \frac{A_n}{A_{n-1}}$$with $A_{-1}=0$, is the associated graded algebra. Let define the map
$$Symb:A \rightarrow Gr(A)$$ that send $x \in A$ in $[x]\in A_n/A_{n-1}$ where $n$ is the minimum $n$ such that $x \in A_n$. I want to find a solution for the following questions:
Symb is not in general linear
If the filtration on $A$ is induced by a graduation, i.e. $A=\bigoplus \bar{A}_n$ and $A_N=\bigoplus_{m=0}^{N}\bar{A}_m$. Then Symb is an isomorphism? A and Gr(A) are isomorphic?
$\operatorname{Symb}$ 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 $\operatorname{Symb}(x^2)+\operatorname{Symb}(x)=x^2+x$.