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Suppose $A$ and $B$ are two filtered associative $k$-algebras with increasing filtrations $F^{bullet}$ and $G^{bullet}$, respectively. It is safe to assume that the filtrations are exhaustive. Assume that there is a filtered algebra morphism $f:Arightarrow B$, that is, f satisfies $f(F^n)subseteq G^n$. If the induced map $gr(f): gr_{F^{bullet}}(A)rightarrow gr_{G^{bullet}}(B)$ is an isomorphism of graded $k$-algebras, is it true that the original map $f$ is also a $k$-algebra isomorphism. If not, under what conditions would that be true? Thank you for your help and feedback.

I think it's true, but you have to assume the filtrations are exhaustive.

I'll use a bar to denote the associated graded, e.g. $bar f = gr(f)$, $bar A^k = F^kA/F^{k-1}A$ and if $y in G^kBsetminus G^{k-1}B$ then $bar y in bar B^k$. (It would have been better notation to have defined $pi_{k}: F^kA to F^kA/F^{k-1}A$ and kept track of $k$.)

Injectivity:
Assume $bar f$ is an isomorphism. Let $x in A$ and take the minimal $k$ such that $x in F^kA$. Then $bar x in bar A^k$ is nonzero, so $bar f(bar x) neq 0$. But $bar f(bar x) = overline{f(x)}$ so this means $f(x) neq 0$.

Note. It's important that $bigcup F^kA = A$, otherwise we can't find a minimal such $k$ (the set of $k$ with $x in F^kA$ may be empty).

Surjectivity:
Let $y in G^kB$ for minimal $k$ and $bar y in bar B^k$ the corresponding element. Take a preimage $bar x in bar A^k$ and lift this to $x in F^kA$. We can't say that $f(x) = y$, but at least $y_1 := y - f(x) in G^{k-1}B$. Now we repeat for $y_1$ and find $x_1$ such that $y_1 - f(x_1) in G^{k-2}B$. And so on until $k-i = 0$ and we can find a preimage $x + x_1 + x_2 + cdots + x_{k-1}$.

Note. It's important that $bigcup G^kB = B$, else we can't find a minimal $k$ again.

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Added:

In the case of decreasing sequences, I think you can prove injectivity the same way but you need to assume $bigcap F^kA=0$. For surjectivity you need to assume completeness with respect to the $G$ filtration so that the infinite sum makes sense and probably also the intersection condition.

Here's another proof, which is basically the same proof as last time, but references the Five lemma.

Lemma. For the map $f: A to B$, if the induced map $$gr^{leq p}_F(A) to gr^{leq p}_G(B)$$ is an isomorphism then $F^iA to F^iB$ is an isomorphism for $i leq p$.

Proof. If $p=0$ the $gr^0_F(A) = F^0A, gr^0_GB = G^0B$ so there is nothing to prove. Now assume $F^{p-1}A to F^{p-1}B$ is an isomorphism. We have a morphism between exact sequences:

$0 to F^{p-1}A to F^pA to gr_F^p(A) to 0$

$0 to G^{p-1}B to G^pB to gr_G^p(B) to 0$

Where all the arrows except the middle we know are isomorphisms. By the five lemma, the middle is an isomorphism, too. $square$

Now we know $F^iA to G^iB$ is an isomorphism for all $i$, and hence that $$bigcup F^iA to bigcup G^iB$$ is an isomorphism. So long as the filtrations are exhaustive, we're done.