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Disclaimer : I asked this question on MSE, I have no answer and I think it's better to ask it here.

Let $(A,\mathcal{W}, \mathcal{C})$ be a Waldhausen category with $A$ an additive category. On one hand, we can define the ordinary limits $lim_A$ of the underlying category $A$. On other hand, we can define limits of Waldhausen categories via the universal property of a diagram with some arrows in $\mathcal{C}$ . For example we can define $ker_{\mathcal{C}}(f) \stackrel{i}{\rightarrow}X\stackrel{f}{\rightarrow}Y$ where $i \in \mathcal{C}$ has the universal property of the kernel for $j \in \mathcal{C} | fj=0$.

My question is: if they exist, do ordinary limits and Waldhausen limits commute? In particular, do ordinary countable products and Waldhausen kernels commute?

Or : do we have some conditions such that ordinary limits and Waldhausen limits commute?

(I'm interested by the second case, but the first one implies the second one. And in my particular case $\mathcal{W}=Iso$)

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Unless I'm misunderstanding your question, the answer is yes. Your second way of defining limits, via "the universal property of a diagram with some arrows in $\mathcal{F}$," is actually a special case of the normal definition of a limit, and limits commute. For your specific question of interest, the key observation is that $ker_{\mathcal{F}}(f)$ is the pullback of the co-span below.

$$ \require{AMScd} \begin{CD} ker_{\mathcal{F}}(f) @>>> X \\ @VVV @VVV \\ 0 @>>> Y \end{CD} $$

where $0$ is the zero object.

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  • $\begingroup$ I edited my answer, I'm looking for Waldhausen categories and not coWaldhausen... Since we don't know that cofibration are closed under pullback, in your diagram we want to have $ker_\mathcal{C}(f) \rightarrow X$ a cofibration. I agree that limits commute but I'm not seeing why $ker_\mathcal{C}$ is a limit in $A$... $\endgroup$ – MoreauT Jun 17 at 13:35
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    $\begingroup$ In my answer, nowhere it is it used that $f: X\to Y$ is a fibration. It works just as well if $f$ is a cofibration instead. The point is that kernel is a pullback, hence a limit, regardless of the (co)fibration structure. You could choose any class of (co)fibrations and kernel would still be a limit. Check out Weibel's "The K Book" to learn more. $\endgroup$ – David White Jun 17 at 15:18
  • $\begingroup$ You seem to claiming that the ker_C = ker ? if I understand the question correctly, ker_C is essentially the kernel in the category "C" (the non full subcategory of all objects and arrows that are in $C$). $\endgroup$ – Simon Henry Jun 19 at 15:01
  • $\begingroup$ Hmm, maybe. He edited the question, and I'm swamped with a July 1 deadline, so not really following the edits. $\endgroup$ – David White Jun 19 at 15:44

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