There is a notion of equivalence of categories which is the functor $F:\mathcal{C} \to \mathcal{D}$ such that there is a functor $G:\mathcal{D} \to \mathcal{C}$ such that $FG \cong id_{\mathcal{D}}$ and $GF \cong id_{\mathcal{C}}$ where $\cong$ means natural isomorphism (natural transformation with each component being the isomorphism). One can also check equivalence of categories without referring to $G$: it suffices that $F$ is full, fauthul (faithful means injective on arrows between fixed objects, full means surjective on arrows between $F(C_1)$ and $F(C_2)$) and essentially surjective (surjective on objects, up to isomorphism). Note that from being full and faithul follows that $F$ is essentially injective on objects (meaning that if $F(C_1)$ and $F(C_2)$ are isomorphic in $\mathcal{D}$ then $C_1$ and $C_2$ are isomorphic in $\mathcal{C}$). Is there an example of two categories $\mathcal{C}$ and $\mathcal{D}$ with the properties:
-there is a functor $F:\mathcal{C} \to \mathcal{D}$ being essential bijection on objects (meaning essentially surjective and essentially injective in the above sense)
-for each two objects $C_1,C_2 \in \mathcal{C}$ there is an abstract bijection between $Hom_{\mathcal{C}}(C_1,C_2) \simeq Hom_{\mathcal{D}}(F(C_1),F(C_2))$ meaning that this bijection is not implemented by $F$.
To be more precise: for each pair of objects $C_1,C_2 \in \mathcal{C}$ we can find a map $T_{C_1,C_2}:Hom_{\mathcal{C}}(C_1,C_2) \to Hom_{\mathcal{D}}(F(C_1),F(C_2))$ being a bijection of sets and this map is not of the form $F(f)$ for $f \in Hom_{\mathcal{C}}(C_1,C_2)$?
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$\begingroup$ What is an "abstract" bijection and what does it mean for "a bijection to be implemented by $F$"? $\endgroup$– Andrej BauerCommented May 15, 2016 at 18:14
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$\begingroup$ 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 $F(f)$... $\endgroup$– truebaranCommented May 15, 2016 at 18:26
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1$\begingroup$ ... this I meant to be implemented by $F$, $\endgroup$– truebaranCommented May 15, 2016 at 18:26
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2$\begingroup$ 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'. $\endgroup$– David Roberts ♦Commented May 15, 2016 at 20:52
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1$\begingroup$ David Roberts thank you, you are obviosly right, I corrected what was wrong $\endgroup$– truebaranCommented May 15, 2016 at 21:20
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1 Answer
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Let $D$ be the category whose only object is $\mathbb{N}$ and the morphisms are all functions $\mathbb{N} \to \mathbb{N}$. Let $C$ be the subcategory whose morphisms are bijections $\mathbb{N} \to \mathbb{N}$.
The inclusion $F : C \to D$ has the properties you ask for.
answered May 15, 2016 at 18:38