Faithful functor

Faithful functors and the full and fully faithful functors to be discussed here , which are closely related to them, are functors with special properties considered in the mathematical theory of category theory .

Definitions

Be a functor between two categories and . Such functor assigns definition, every object and every morphism from where and objects were, an object or a morphism to, certain compatibility conditions are met. ${\displaystyle T\colon {\mathcal {C}}\rightarrow {\mathcal {D}}}$ ${\displaystyle {\mathcal {C}}}$ ${\displaystyle {\mathcal {D}}}$ ${\displaystyle X\in \operatorname {Ob} ({\mathcal {C}})}$ ${\displaystyle f\colon X\rightarrow Y}$ ${\displaystyle \operatorname {Mor} _{\mathcal {C}}(X,Y)}$ ${\displaystyle X}$ ${\displaystyle Y}$ ${\displaystyle {\mathcal {C}}}$ ${\displaystyle T(X)\in \operatorname {Ob} ({\mathcal {D}})}$ ${\displaystyle T(f)\in \operatorname {Mor} _{\mathcal {D}}(T(X),T(Y))}$

For each pair of objects (in the case of locally small categories) one has a mapping ${\displaystyle (X,Y)}$ ${\displaystyle {\mathcal {C}}}$

${\displaystyle T_{X,Y}\colon \operatorname {Mor} _{\mathcal {C}}(X,Y)\,\rightarrow \,\operatorname {Mor} _{\mathcal {D}}(T(X),T(Y)),\,f\mapsto T(f).}$

The functor is called true (or fully or fully faithful ) if the mappings for each pair of objects are injective (or surjective or bijective ). Instead of fully faithful , the term fully faithful is also found . ${\displaystyle T}$ ${\displaystyle T_{X,Y}}$ ${\displaystyle (X,Y)}$ ${\displaystyle {\mathcal {C}}}$

Embeddings

If a functor is, the terms true , full and completely true only refer to sets of morphisms between two objects, they do not refer to the classes of all objects or all morphisms, in particular the fidelity of the functor does not necessarily mean that one of the mappings ${\displaystyle T\colon {\mathcal {C}}\rightarrow {\mathcal {D}}}$ ${\displaystyle T}$

${\displaystyle T_{\text{Ob}}\colon \operatorname {Ob} ({\mathcal {C}})\rightarrow \operatorname {Ob} ({\mathcal {D}}),\,X\mapsto T(X),}$

${\displaystyle T_{\text{Mor}}\colon \operatorname {Mor} ({\mathcal {C}})\rightarrow \operatorname {Mor} ({\mathcal {D}}),\,f\mapsto T(f)}$

is injective. In order to illuminate the context of these terms and the use of the above definitions, the following simple statement is proven here:

• If the functor is faithful, then is injective if and only if is injective. ${\displaystyle T}$ ${\displaystyle T_{\text{Ob}}}$ ${\displaystyle T_{\text{Mor}}}$

Is injective and are with , it follows , that is, according to the assumption and with it . Hence is injective. ${\displaystyle T_{\text{Mor}}}$ ${\displaystyle X,Y\in \operatorname {Ob} ({\mathcal {C}})}$ ${\displaystyle T(X)=T(Y)}$ ${\displaystyle T(\operatorname {id} _{X})=\operatorname {id} _{T(X)}=\operatorname {id} _{T(Y)}=T(\operatorname {id} _{Y})}$ ${\displaystyle \operatorname {id} _{X}=\operatorname {id} _{Y}}$ ${\displaystyle X=Y}$ ${\displaystyle T_{\text{Ob}}}$

Now, conversely, be injective, and be with . It is to be shown. The morphisms and include objects from the category with and . It follows from and . Because by assumption is injective, we get and . Therefore, and the loyalty of delivers as desired . ${\displaystyle T_{\text{Ob}}}$ ${\displaystyle f,g\in \operatorname {Mor} ({\mathcal {C}})}$ ${\displaystyle T(f)=T(g)}$ ${\displaystyle f=g}$ ${\displaystyle f}$ ${\displaystyle g}$ ${\displaystyle X_{1},X_{2},Y_{1},Y_{2}}$ ${\displaystyle {\mathcal {C}}}$ ${\displaystyle f\colon X_{1}\rightarrow Y_{1}}$ ${\displaystyle g\colon X_{2}\rightarrow Y_{2}}$ ${\displaystyle T(f)=T(g)}$ ${\displaystyle T(X_{1})=T(X_{2})}$ ${\displaystyle T(Y_{1})=T(Y_{2})}$ ${\displaystyle T_{\text{Ob}}}$ ${\displaystyle X_{1}=X_{2}}$ ${\displaystyle Y_{1}=Y_{2}}$ ${\displaystyle T_{X_{1},Y_{1}}(f)=Tf=Tg=T_{X_{1},Y_{1}}(g)}$ ${\displaystyle T}$ ${\displaystyle f=g}$

A functor is called an embedding if it is injective. For a true functor, the embedding property according to the above is equivalent to the injectivity of . ${\displaystyle T}$ ${\displaystyle T_{\text{Mor}}}$ ${\displaystyle T_{\text{Ob}}}$

If the functor is an embedding, the objects with the morphisms form a subcategory of , which is denoted by. Since this is generally not the case for arbitrary functors that are not embedded, embeddings play an important role in category theory. ${\displaystyle T\colon {\mathcal {C}}\rightarrow {\mathcal {D}}}$ ${\displaystyle T(X),X\in \operatorname {Ob} ({\mathcal {C}})}$ ${\displaystyle T(f),f\in \operatorname {Mor} ({\mathcal {C}})}$ ${\displaystyle {\mathcal {D}}}$ ${\displaystyle T({\mathcal {C}})}$

Fully faithful functors

If the functor is an embedding, and is a full functor, then is a full subcategory of . This motivates the term full functor in the above definitions. So is a fully true functor, so that is injective, then defines an embedding on a full subcategory. ${\displaystyle T\colon {\mathcal {C}}\rightarrow {\mathcal {D}}}$ ${\displaystyle T}$ ${\displaystyle T({\mathcal {C}})}$ ${\displaystyle {\mathcal {D}}}$ ${\displaystyle T}$ ${\displaystyle T_{\text{Ob}}}$ ${\displaystyle T}$

Fully true functors are also important for category theory because of the following statement:

• Be a fully faithful functor and category morphism . Then: is isomorphism is isomorphism. ${\displaystyle T\colon {\mathcal {C}}\rightarrow {\mathcal {D}}}$ ${\displaystyle f\colon X\rightarrow Y}$ ${\displaystyle {\mathcal {C}}}$ ${\displaystyle f}$ ${\displaystyle \Leftrightarrow }$ ${\displaystyle Tf}$

The left to right direction is very easy. If namely isomorphism, there is by definition another morphism with and . Since functor is, it follows and likewise , that is, is an isomorphism. ${\displaystyle f}$ ${\displaystyle g\colon Y\rightarrow X}$ ${\displaystyle fg=\operatorname {id} _{Y}}$ ${\displaystyle gf=\operatorname {id} _{X}}$ ${\displaystyle T}$ ${\displaystyle T(f)\circ T(g)=T(fg)=T(\operatorname {id} _{Y})=\operatorname {id} _{T(Y)}}$ ${\displaystyle T(g)\circ T(f)=\operatorname {id} _{T(X)}}$ ${\displaystyle T(f)}$

Full fidelity is required for the reversal. For if there is an isomorphism, there is a morphism with and . Since it is full, there is a morphism with . Then follows and exactly the same . Because of the fidelity of , it now follows and , that is, is an isomorphism. ${\displaystyle T(f)\colon T(X)\rightarrow T(Y)}$ ${\displaystyle w\colon T(Y)\rightarrow T(X)}$ ${\displaystyle T(f)\circ w=\operatorname {id} _{T(Y)}}$ ${\displaystyle w\circ T(f)=\operatorname {id} _{T(X)}}$ ${\displaystyle T}$ ${\displaystyle g\colon Y\rightarrow X}$ ${\displaystyle T(g)=w}$ ${\displaystyle T(fg)=T(f)\circ T(g)=T(f)\circ w=\operatorname {id} _{T(Y)}=T(\operatorname {id} _{Y})}$ ${\displaystyle T(gf)=T(\operatorname {id} _{X})}$ ${\displaystyle T}$ ${\displaystyle fg=\operatorname {id} _{Y}}$ ${\displaystyle gf=\operatorname {id} _{X}}$ ${\displaystyle f}$

literature

• Horst Schubert : Categories (= Heidelberger Taschenbücher. 65–66, ISSN  0073-1684 ). Volume 1-2. Springer, Berlin et al. 1970.