I have a question about the equivalence of derived categories. Let $\mathcal{A} = \mathcal{A}'\oplus \mathcal{A}''$ and $\mathcal{B} = \mathcal{B}' \oplus \mathcal{B}''$ are direct sum of abelian categories. Let $T$ be an complex of exact functors $$T: \text{ }\cdots \rightarrow T^{p-1} \rightarrow T^{p} \rightarrow T^{p+1} \rightarrow \cdots$$ (each $T^q$ is an functor from $\mathcal{A}$ to $\mathcal{B}$). Assume that for $T(X)$ is bounded for each $X\in \mathcal{A}$, then we can obtain a functor between their bounded derived categories, i.e. $T:$ $\mathcal{D}^b(\mathcal{A}) \rightarrow \mathcal{D}^b(\mathcal{B})$, by taking total complexes.

Assume that $T$ is an equivalence as a triangulated equivalence. Then it induce a isomorphism $[T]$ between their Grothendieck groups $[T]:\mathcal{K}(\mathcal{A}) \rightarrow \mathcal{K}(\mathcal{B})$. (Note that $\mathcal{K}(\mathcal{\mathcal{C}}) \cong \mathcal{K}(\mathcal{D}^b(\mathcal{C}) ) $ for every abelian category $\mathcal{C}$).

$\bf My$ $\bf Question:$ Is it true that $[T](\mathcal{K}(\mathcal{A}')) = \mathcal{K}(\mathcal{B}')$ implies the restriction $T|_{\mathcal{A}'}$ to $\mathcal{A}'$ is an equivalence between $\mathcal{D}^b(\mathcal{A'}) $ and $\mathcal{D}^b(\mathcal{B'})$? Furthermore, I was wondering when this is true. Thanks very much!