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. $l\circ l = l$ where $l=u\circ f$.

3. Let $b \in B $, if $f(b)=0$ then $b=0$.

Question: do we have $K_{0} (A)= K_{0}(B)$ ?

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 $K_{0} (A)= K_{0}(B)$ ?

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. $l\circ l = l$ where $l=u\circ f$.

3. Let $b \in B $, if $f(b)=0$ then $b=0$.

Question do we have that $K_{0} (A)= K_{0}(B)$ ?

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. $l\circ l = l$ where $l=u\circ f$.

3. Let $b \in B $, if $f(b)=0$ then $b=0$.

Question: do we have $K_{0} (A)= K_{0}(B)$ ?