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Aug 27, 2021 at 16:37 vote accept Ho Man-Ho
Aug 27, 2021 at 16:07 answer added Dmitri Pavlov timeline score: 2
Aug 27, 2021 at 15:59 comment added Ho Man-Ho @AchimKrause If I say for any two manifolds M and N and a smooth map $f:N\to M$ (which is a morphism), $\Omega(f):\Omega(M)\to\Omega(N)$ is the pullback of forms, does it make sense?
Aug 27, 2021 at 15:55 comment added Achim Krause Superficially, it looks like you want a natural transformation between functors $\operatorname{Man} \to \operatorname{Cat}$. However, for that to make sense, you need to make sense of $\Omega(M)$ as a category, and pointwise $T(M): \operatorname{Vect}_\Delta(M) \to \Omega(M)$ as a functor of categories.
Aug 27, 2021 at 15:36 history edited Ho Man-Ho CC BY-SA 4.0
Thanks to AT0 for correcting a mistake.
Aug 27, 2021 at 15:31 comment added Ho Man-Ho I am probably wrong. In principle $\Omega^*:\textrm{Man}^{\textrm{op}}\to\textrm{Set}$ should be a functor to the category of sets. I will revise my question. Thank you.
Aug 27, 2021 at 13:29 comment added AT0 I dont understand what is 'the category of differential forms on M', what category structure are you considering?
Aug 27, 2021 at 12:38 history edited Ho Man-Ho CC BY-SA 4.0
added 2 characters in body
Aug 27, 2021 at 12:35 comment added Ho Man-Ho Thanks, I will check if 2-categories suffice
Aug 27, 2021 at 11:10 comment added Fernando Muro I think 2-categories would suffice here, but I may be wrong
Aug 27, 2021 at 9:48 history asked Ho Man-Ho CC BY-SA 4.0