Let $A$ be a full triangulated subcategory of $B$, $u:A\rightarrow B$ the corresponding embedding. Let $f:B\rightarrow A$ be a triangulated functor
satisfying:


1. $f\circ u = id$
2. Let $b \in B $, if $f(b)=0$ then $b=0$.


**Question:** do we have $K\_{0} (A)= K\_{0}(B)$ ?



Grothendieck group of triangulated categories
Let $A$ be a triangulated category, and let $B=A\times A$, with $A$ regarded as a full triangulated subcategory of $B$ via the embedding $u(X)=(X,0)$, and let $f:B\to A$ be the functor $f(X,Y)=X\oplus Y$.


Then $f\circ u=\text{id}\_A$, and $K\_0(B)\cong K\_0(A)\oplus K\_0(A)$, which might not be isomorphic to $K\_0(A)$.


