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May 16, 2016 at 6:28 review Reopen votes
May 16, 2016 at 8:04
May 16, 2016 at 6:11 history edited truebaran CC BY-SA 3.0
added 283 characters in body; edited title
May 15, 2016 at 23:45 history closed Andrej Bauer
abx
Qiaochu Yuan
Wolfgang
Myshkin
Needs details or clarity
May 15, 2016 at 21:20 comment added truebaran David Roberts thank you, you are obviosly right, I corrected what was wrong
May 15, 2016 at 21:18 history edited truebaran CC BY-SA 3.0
added 86 characters in body
May 15, 2016 at 20:52 comment added David Roberts It is false that faithful means "injective on arrows". Please check the definition again, as the real definition of fully faithful doesn't imply injectivity on objects. Eg the functor M from the category of vector spaces over R with basis, and all linear maps, to the category with objects the natural numbers and matrices as morphisms: this is an equivalence of categories and highly non-injective on objects and arrows. Also, use 'injective up to isomorphism' becasue that is what you really mean, or 'essentially injective'.
May 15, 2016 at 18:41 vote accept truebaran
May 15, 2016 at 18:38 comment added Andrej Bauer Ok, not really a research-level question in my opinion, it would be better to ask this on math.stackexchange.com.
May 15, 2016 at 18:38 answer added Andrej Bauer timeline score: 7
May 15, 2016 at 18:31 review Close votes
May 15, 2016 at 23:45
May 15, 2016 at 18:26 comment added truebaran ... this I meant to be implemented by $F$,
May 15, 2016 at 18:26 comment added truebaran Not implemented by $F$ means that for each pair of objects $C_1,C_2$ there is a map $T_{C_1,C_2}$ between sets $Hom_{mathcal{C}}(C_1,C_2)$ and $Hom_{\mathcal{D}}(F(C_1),F(C_2))$ which is a bijection. This map sends a morphism $f:C_1 \to C_2$ to some morfizm between $F(C_1) \to F(C_2)$ in such a way that $T_{C_1,C_2}f_1=T_{C_1,C_2}f_2$ implies $f_1=f_2$ and each morphism in $Hom_{\mathcal{D}}(F(C_1),F(C_2))$ is of the form $T_{C_1,C_2}(f)$ for some morphism $f \in Hom_{\mathcal{C}}(C_1,C_2)$. In the definition of equivalence of categories $T_{C_1,C_2}(f)$ is just $F(f)$...
May 15, 2016 at 18:14 comment added Andrej Bauer What is an "abstract" bijection and what does it mean for "a bijection to be implemented by $F$"?
May 15, 2016 at 18:05 history asked truebaran CC BY-SA 3.0