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Jan 19, 2023 at 7:19 comment added naahiv Regarding Martin's third example and making a choice $A\to I$ natural in $A$: since, as you said, the $I\to I$ component should be an isomorphism, any $f,g\colon I\to A$ are necessarily equal by naturality, and thus $I$ is initial in $\mathcal{C}$. Therefore the construction works for semicocartesian categories (with a sequentially cocontinuous product). Just naming what you already wrote for future readers.
Aug 21, 2013 at 22:50 comment added fosco I want to publicly thank you for answering to (part of) this related question mathoverflow.net/questions/138518/…
Sep 12, 2012 at 9:05 comment added Zhen Lin I'm not sure how to resolve that problem. Perhaps zero should be excluded everywhere – leading to some kind of non-unital infinitary monoidal category – and then we adjoin the axioms for the unit from the theory for finitary monoidal categories. Just excluding partitions containing infinitely many copies of $0$ is troublesome for the statement of the associative law: if we do that, then $1 + 1 + 1 + \cdots = \omega$ and $0 + 1 = 1$ are both valid partitions, but the substituted partition $(0 + 1) + (0 + 1) + (0 + 1) + \cdots = \omega$ is not.
Sep 12, 2012 at 7:40 comment added Martin Brandenburg Thank you Zhen. But this does not include the tensor product of modules, as you say. If $K$ is a field, then $\bigotimes_{i < \alpha} K$ is a twisted group algebra of the giant group ${K^*}^{\alpha} / {K^*}^{({\alpha})}$ (see the link in my question). So this contradicts the definition with the trivial partition $0 = \sum_{i < \alpha} 0$. How can we adjust the definition so that it works? For example, we may treat the zero case separatedly.
Sep 8, 2012 at 20:07 history edited Zhen Lin CC BY-SA 3.0
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Sep 8, 2012 at 19:23 history edited Zhen Lin CC BY-SA 3.0
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Sep 8, 2012 at 19:17 history answered Zhen Lin CC BY-SA 3.0