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Jan 16, 2018 at 8:01 vote accept Abo Kutis-Felan
Jan 16, 2018 at 7:58 comment added Abo Kutis-Felan Yes, I guess additivity should be assumed, and then the condition is as you put it.
Jan 15, 2018 at 23:07 answer added Noah Snyder timeline score: 1
Jan 15, 2018 at 20:35 comment added Tobias Fritz Ah, okay! So then you probably want $f$ to be additive as well? If so, then your property is equivalent to requiring $f(A\otimes B)\cong f(A)\otimes f(B)$ for all objects $A$ and $B$, but without postulating any coherences, right?
Jan 15, 2018 at 19:32 comment added Abo Kutis-Felan At Tobias: Sorry, there was a typo in question. It's now fixed, and the property in non longer implied by additivity.
Jan 15, 2018 at 19:31 history edited Abo Kutis-Felan CC BY-SA 3.0
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Jan 15, 2018 at 18:20 comment added Tobias Fritz Trivially every additive $f$ has that property. Do you know of any non-additive functor for which this holds?
Jan 15, 2018 at 17:15 history asked Abo Kutis-Felan CC BY-SA 3.0