Isomorphism between a filtered vector space and its associated graded $\DeclareMathOperator\gr{gr}$Let $ V $ be a vector space with a decreasing filtration
$$
V = F_0 V \supseteq F_1 V \supseteq F_2 V \supseteq\dotsb .$$
We define the associated graded of $ V $ to be $$ \gr V := \bigoplus_{k=0}^\infty F_k V / F_{k+1} V. $$
Of course $ \gr V $ can also be regarded as a filtered vector space and we have a canonical isomorphism $\gr (\gr V) = \gr V $.
We say that $ V $ “admits an expansion” if there is an isomorphism of filtered vector spaces between $ \gr V $ and $ V $, which becomes the identity map after applying $ \gr $ to both $ \gr V $ and $ V $.
This condition is equivalent to the existence of subspaces $ W_k \subset F_k V $ such that $ F_k V = W_k \oplus F_{k+1} V $ and $ V = \bigoplus_k W_k $.
Note that not every filtered vector space admits an expansion.  For example, the vector space $ V = \mathbb C[[x]] $ with the filtration $ F_k V = x^k \mathbb C[[x]]$ does not admit an expansion.  On the other hand, $ V = \mathbb C[x] $ with the same filtration does admit an expansion.
Here are my questions:

*

*Does this property have a different name in the literature?

*Let $V$, $W $ be two filtered vector spaces which admit expansions.  Suppose that I have a filtration-preserving map $ \phi : V \rightarrow W $ such that $ \gr \phi : \gr V \rightarrow \gr W $ is an isomorphism.  Can I conclude that $ \phi $ is an isomorphism?

 A: Too long for a comment, I was wondering about the cost of completely unfolding Darij's excellent argument. In fact, we have a characterisation of that sort of perturbations of identity that are isomorphisms. The statement is as follows 
Lemma. Let $k$ be a field and $x$ an indeterminate. For every $Q\in k[x]$, let $f_Q$ be the morphism of $k[x]$ sending $x$ to $x+x^2Q$. Then $f_Q$ is an isomorphism iff $Q=0$.
Proof. One way being obvious (if $Q=0$ $f_Q=Id$), it is sufficient to prove that, if $Q\not=0$ then
$x\notin Im(f_Q)=f_Q(k[x])$.
It is not difficult to see that $Im(f_Q)$ is a subalgebra as follows
$$Im(f_Q)\subset k\oplus (x+x^2Q)k[x]$$
which does not contain $x$ (indeed $P\in Im(f_Q)$ implies that $P(x)-P(0)$ can be divided by
$(x+x^2Q)$ which is true for $P=x$ iff $Q=0$).
A: 

*NO. take V=W. If the map is identity on the grading, it only means the diagonal blocks are each identity. But the map only needs to be lower triangular. Restriction on the diagonal blocks leaves a lot of freedom for the map.

