Let $\mathcal{C}$ be a Waldhausen category. Suppose that $\mathcal{B}$ is a subcategory of $\mathcal{C}$, and that $\mathcal{B}$ is closed under extensions. If $\mathcal{B}$ is strictly cofinal in $\mathcal{C}$ (in the sense that given any $C\in \mathcal{C}$ there exists a $B\in \mathcal{B}$ such that $C\amalg B\in \mathcal{B}$), can we say anything about $K(\mathcal{B}) \rightarrow K(\mathcal{C})$?

In Waldhausen's paper "Algebraic K-theory of spaces" Waldhausen claims that the inclusion $\mathcal{B}\rightarrow \mathcal{C}$ induces a weak equivalence $wS_\bullet \mathcal{B}\rightarrow wS_\bullet\mathcal{C}$ (and thus an equivalence on K-theories), but I'm not sure that this is right, as $\mathcal{B}$ does not need to be a full subcategory. In particular, if there are objects $C,C'$ which are in $\mathcal{B}$ but are not isomorphic in $\mathcal{B}$ they may well be isomorphic (or at least weakly equivalent) in $\mathcal{C}$.

Consider the following example. Let $\mathcal{C}$ be the category of pairs of pointed finite sets, whose morphisms $(A,B)\rightarrow (A',B')$ are pointed maps $A\vee B\rightarrow A'\vee B'$, and let $\mathcal{B}$ be the category of pairs of pointed finite sets whose morphisms $(A,B)\rightarrow (A',B')$ are pairs of pointed maps $A\rightarrow B$ and $A'\rightarrow B'$. We make $\mathcal{C}$ a Waldhausen category by defining the weak equivalences to be the isomorphisms, and the cofibrations to be the injective maps. $\mathcal{B}$ is clearly cofinal in $\mathcal{C}$, but $K_0(\mathcal{B}) = \mathbf{Z}\times \mathbf{Z}$, while $K_0(\mathcal{C}) = \mathbf{Z}$. Going even further, the Barratt-Priddy-Quillen theorem should tell us that $K(\mathcal{B}) = QS^0\times QS^0$, while $K(\mathcal{C}) = QS^0$.

If we add the condition that $\mathcal{B}$ needs to be a full subcategory of $\mathcal{C}$, then I believe that Waldhausen's paper is correct. But even without that, it is possible to say anything about the map $K(\mathcal{B})\rightarrow K(\mathcal{C})$?