CatDat

Implication Details

Assumptions: left-invertible

Conclusions: conservativeessentially injectivefaithful

Proof: Let G:DCG : \D \to \C be a left-inverse to F:CDF : \C \to \D, meaning that GFidCG \circ F \cong \id_{\C}. Then F(A)F(B)F(A) \cong F(B) implies AG(F(A))G(F(B))BA \cong G(F(A)) \cong G(F(B)) \cong B for all A,BCA,B \in \C. Thus, FF essentially injective. Moreover, since GFG \circ F is faithful, the composed map Hom(A,B)Hom(F(A),F(B))Hom(G(F(A)),G(F(B))\Hom(A,B) \to \Hom(F(A),F(B)) \to \Hom(G(F(A)),G(F(B)) is injective, so that also Hom(A,B)Hom(F(A),F(B))\Hom(A,B) \to \Hom(F(A),F(B)) is injective. This shows that FF is faithful. Finally, if f:ABf : A \to B is am morphism such that F(f)F(f) is an isomorphism, then G(F(f))G(F(f)) is an isomorphism. Since G(F(f))fG(F(f)) \cong f in Mor(C)\Mor(\C), we conclude that ff is an isomorphism. Therefore, FF is conservative.

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