Timeline for Why pullback only defined up-to-isomorphism but nevertheless presented as functor?

7 events

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Dec 13, 2012 at 6:30	comment	added	Almeo Maus		Every answer brought me useful insight, so it is difficult to decide which is the "good one". However the answer of Andrej pointed that what I am stuck upon is the same notion that gives a product of two objects as an "operation" even when there are infinities of "product candidates", by taking one as a "reference" and calling it "canonical" then calling the others "canonically isomorphic" and the isomorphisms that link them to the canonical one "canonical isomorphisms". That does not answer all the pb but answers from Todd and David complete it. Moreover Andrej has published with Awodey :)
Dec 13, 2012 at 6:16	vote	accept	Almeo Maus
Dec 12, 2012 at 0:59	history	edited	Andrej Bauer	CC BY-SA 3.0	added 47 characters in body
Dec 11, 2012 at 15:53	vote	accept	Almeo Maus
Dec 11, 2012 at 16:44
Dec 11, 2012 at 15:12	vote	accept	Almeo Maus
Dec 11, 2012 at 15:38
Dec 11, 2012 at 15:00	comment	added	Almeo Maus		Thank you very much for your answer !! Yes, you must be speaking about the functorial relation on compositions, aren't you? By the way, is that the relation that does not hold when we add fancy isomorphisms to all the $\alpha'$, as you replied to Zhen and Steven? So, any family of couples $(\alpha,\alpha')$ so that the square is always a pullback and so that the composition condition for functors is satisfied, is a definition of a functor for this pullback?
Dec 11, 2012 at 14:17	history	answered	Andrej Bauer	CC BY-SA 3.0