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I was recently thinking about the proof of a theorem where Waldhausen compared the Segal's delooping machinery with his, in the case when the cofibration is splittable (sec.1.8 in 'Algebraic $K$-theory of space').

Assume $\mathcal{C}$ is a Waldhausen category. By forgetting some structures, one obtains a category with sum (and weak equivalence). We denote by $wN_{\cdot}$ the Segal's delooping machinery developed in `Categories and cohomology theories´

Now the comparison map is
$$wN_{\cdot}\mathcal{C}\rightarrow wS_{\cdot}\mathcal{C},$$ and by delooping them once, he turned the question into proving $$wN_{\cdot}N_{\cdot}\mathcal{C}\rightarrow wN_{\cdot}S_{\cdot}\mathcal{C}$$ is a homotopy equivalence. Then he purposed to prove an additive theorem for $N_{\cdot}$-construction, namely, to show $$wN_{\cdot}S_{n}\mathcal{C}\rightarrow (wN_{\cdot}\mathcal{C})^{n}$$ is a homotopy equivalence (Prop. 1.8.7). It is at this stage in his proof the splittabe condition is made use of. However, instead of using the approach in his proof of the additive theorem for $S_{\cdot}$- construction, he used a quite different way to show this.(see Remark below)

My question or confusion is that: Can we still imitate the method he used in the proof of the additive theorem for $S_{\cdot}$-construction and apply it to the $N_{\cdot}$-construction, meaning, we first try to prove

$$ wN_{\cdot}E(\mathcal{C})\rightarrow wN_{\cdot}\mathcal{C}\times wN_{\cdot}\mathcal{C}$$

is a homotopy equivalence, where $E(\mathcal{C})$ is the category of cofibration sequence or $S_{2}\mathcal{C}$. And then by the other equivalent statements of the additive theorem, one can obtain the homotopy equivalence $$wN_{\cdot}S_{n}\mathcal{C}\rightarrow (wN_{\cdot}\mathcal{C})^{n}.$$

I saw no obvious obstruction. For example, one can replace $s_{k}\mathcal{C}\equiv Obj(S_{k}\mathcal{C})$ by $n_{k}(\mathcal{C})\equiv Obj(N_{k}\mathcal{C})$ and $E(S_{n}\mathcal{C})$ by $E(N_{n}\mathcal{C})$, and then proceed with his argument in sec. 1.4 (the additive theorem) in the same article. But if it is the case, then we don't need splittable condition any more. I wonder where I get wrong. It seems to me very strange. They shouldn't be identical without splittable condition, right?

Remark: Waldhausen in his article used a quite different approach to prove the statement: $$wN_{\cdot}S_{n}\mathcal{C}\rightarrow wN_{\cdot}S_{n-1}\mathcal {C}\times wN_{\cdot}\mathcal{C}$$ is a homotopy equivalence. He first used the homotopy fibration sequence $$wN_{\cdot}\mathcal{D}\rightarrow wN_{\cdot}N_{\cdot}(\mathcal{C}\rightarrow \mathcal{D})\rightarrow wN_{\cdot}N_{\cdot}\mathcal{D}$$ to turn the question into showing $$wN_{\cdot}(j_{n}:\mathcal{C}\rightarrow S_{n}\mathcal{C})\rightarrow wS_{n-1}\mathcal {C}$$ is a homotopy equivalence. Then by studying its fiber, he concluded it is a homotopy equivalence when the cofibration is splittable.

I appreciate any comment or suggestion, and thank you in advance.

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There is no hope of showing that $w N. E(\mathcal C) \to w N. \mathcal C \times w N. \mathcal C$ is a homotopy equivalence without assuming that the cofibrations of $\mathcal C$ are splittable. To see that, take a simple example and look at the direct sum Grothendieck groups which appear as $\pi_1$ of these K-theory spaces. Let $\mathcal C$ be the category of finitely generated abelian groups with the non-split short exact sequence $M \in E(\mathcal C)$ given by $0 \to \mathbb Z \to \mathbb Z \to \mathbb Z/2 \to 0$, and consider the class $[M]$ in $K_0^\oplus \mathcal C$. From an equation $[M]=[M']$, where $M'$ is a splittable short exact sequence, we can deduce that $M$ and $M'$ are stably isomorphic, i.e., there is another short exact sequence $N$ with $M \oplus N \cong M' \oplus N$. The functor $E( \mathcal C ) \to \mathcal C$ that computes the kernel of the tensor product with $\mathbb Z/2$ of the first map in the short exact sequence is an additive functor that sends $M$ to $\mathbb Z/2$, $M'$ to zero, and $N$ to a finite abelian 2-group, yielding an isomorphism between two finite abelian groups of different order, giving a contradiction.

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  • $\begingroup$ Thank you very much! The example is very helpful, and it clarifies my questions! $\endgroup$ – yisheng Dec 16 '15 at 7:55

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