# Do limits in Waldhausen categories commute with ordinary limits?

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 $$\mathcal{W}=Iso$$)

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.
• 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 $A$... – MoreauT Jun 17 at 13:35
• 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. – David White Jun 17 at 15:18
• 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 $C$). – Simon Henry Jun 19 at 15:01