TITLE: L-theory of additive category
QUESTION [5 upvotes]: Reading some articles in the field, I found the following statement:
Proposition:
Let $\mathcal{B}$ be an additive category and $\mathcal{A}$ a full additive subcategory of
$\mathcal{B}$. If $\mathcal{A}\subset \mathcal{B}$ induces an isomorphism in $K$-theory
$$K_{n}(\mathcal{A})\rightarrow_{\cong}K_{n}(\mathcal{B}) $$ for all $n\in \mathbf{Z}$,
then $$L_{n}(\mathcal{A})\rightarrow_{\cong}L_{n}(\mathcal{B}) $$
for all $n\in \mathbf{Z}$.
Is there a proof of such statement?
The proposition  in question is lemma 4.17 of the article:
in their proof the authors say that under the hypothesis of lemma 4.17, any chain complex in $\mathcal{B}$ is homotopy equivalent to a chain complex in $\mathcal{A}$! My question is why it is true?
Here is the exact statement of the lemma:
Lemma 4.17. Let $\mathcal{B}$ be an additive category. If $\mathcal{A}$ is a full subcategory, inducing isomorphism
on $K_{0}$, then
$L(\mathcal{A}) \rightarrow L(\mathcal{B})$
is a weak homotopy equivalence (any decoration on the $L$-theory).

Proof. We need to show the map induces isomorphism on homotopy groups, but any chain complex in $\mathcal{B}$ is homotopy equivalent
to a chain complex in $\mathcal{A}$, so from this it follows that
$L^{h}(\mathcal{A})\rightarrow L^{h}(\mathcal{B})$  is an isomorphism,
and since the inclusion induces isomorphism on $K$-theory Lemma 4.15
finishes off the proof.

I do not understand the meaning of the following argument in their proof :

"but any chain complex in $\mathcal{B}$ is homotopy equivalent to a
chain complex in $\mathcal{A}$".

How do they make this statement in such general stituation. It seems to me that Achim Krause has given a convincing counterexample in his answer. So I'm wondering if I'm missing something in the statement of Lemma 4.17 ?

REPLY [2 votes]: I did find a reference. Sort of. I think the way you stated it, the statement  (I'm referring to the statement about complexes in $\mathcal{B}$ and $\mathcal{A}$, not about the $L$-theory statement) is incorrect. For example, there are rings such that $R\cong R\oplus R$ as modules. For such a ring, $K(R)\simeq 0$, so if you take $\mathcal{B}=\operatorname{Mod}(R)$ and $\mathcal{A}=0$, the map on $K$-theory is an isomorphism, but certainly not every complex in $\mathcal{B}$ is exact.
I believe you additionally need that $\mathcal{A}$ has enough objects, in the sense that every object of $\mathcal{B}$ admits an epimorphism from an object of $\mathcal{A}$ (and possibly more, I'm rarely thinking about bare additive categories). Under the required technical assumptions, you should be able to apply Lemma 2.2 from [Thomason, "The classification of triangulated subcategories", Compositio Mathematica 105: 1–27, 1997] (to the derived categories of $\mathcal{A}$ and $\mathcal{B}$).