Notations : $R$ is a commutative ring with unity. $P(R)$ is the category of finitely generated projective $R-$ modules, $Ch^{b}(P(R))$ is the the category of bounded chain complexes on $P(R)$ and $C^q(P(R))$ is the category of bounded exact chain-complexes on $P(R)$.

Each of the above mentioned categories are exact categories. If I define the weak equivalence class as the isomorphism classes then $K_0$ of each of the categories will be the quotient of the free group generated by the isomorphism classes of the elements $[C]$ where $C \in ob\;\mathcal{C} .$ ($\mathcal{C}$ being any of the above categories). **If we replace $R$ by a field $\mathbb{F}$ then my question is will the inclusion functor $i : C^q(P(\mathbb{F})) \longrightarrow Ch^{b}(P(\mathbb{F}))$ induce an injective group homomorphism from $$K_0C^q(P(\mathbb{F})) \longrightarrow K_0Ch^{b}(P(\mathbb{F}))?$$**
My attempt :

Proposition : For an exact category $\mathcal{C}$ if $[A_1] = [A_2]$ in $K_0(\mathcal{A})$ then there are short exact sequences $0 \rightarrow C' \rightarrow C_1 \rightarrow C'' \rightarrow 0$ and $0 \rightarrow C' \rightarrow C_2 \rightarrow C'' \rightarrow 0$ such that $A_1 \oplus C_1 \cong A_2 \oplus C_2$.

So what I was attempting to show that the kernel of the induced map is trivial, now any typical element in $K_0C^q(P(\mathbb{F}))$ is either $[F.,d]$ or $[F.,d] -[F'.,d']$; ($d,d'$ being the differential). If $[F.,d] \longmapsto 0$ then $F.$ as a complex is itself $0$ according to the proposition (because here the SES of the proposition will split). Thus no problem here

For the second case what I have figured out so far is that if $[F. d] = [F'.,d']$ in $K_0Ch^{b}(P(\mathbb{F}))$ then for each $n$ I have $F_n = F'_n$. (Because there will be a splitting in the SES of the proposition) but then I am unable to proceed further, if you could please point me to the right direction I will be grateful.