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2.Full, Faithful and Embeddings



1. Definition
@2. Full, Faithful and Embeddings
3. Preserves and Reflects
4. Contra-, Co- and Bivariant
5. Co- and Contravariant Hom Functors

Given two categories $\mathcal{C},\mathcal{D}$, and a functor $F : \mathcal{C} \rightarrow \mathcal{D}$, we say that $F$ is

  • full if for every two objects, $A,B$ of $\mathcal{C}$ we have that $$ F : \mathcal{C}(A,B) \rightarrow \mathcal{D}(F(A),F(B)) $$ is a surjection. We can make an example where this is not the case. Denote by subscript the category some object belong to, eg. $$ F(A_C) = A_D $$ Now say we have the arrows $$ f_1,f_2 : A_C \rightarrow B_C $$ in the category $C$. But in the category $D$ we have $$ g_1,g_2,g_3 : A_D \rightarrow B_D $$
  • faithfull if $F$ is a always injective. We can again make a counter example. Say we in the category $\mathcal{C}$ have the arrows $$ f_1,f_2,f_3 : A_C \rightarrow B_C $$ and in the category $D$ we have the arrows $$ g_1,g_2 : A_D \rightarrow B_D $$ Here we let the functor $F$ map as follows $$ [f_1 \mapsto g_1, f_2 \mapsto g_2,f_3 \mapsto g_2] $$

Remeber that $\mathcal{C}(A,B)$ is the class of arrows in $\mathcal{C}$ from the object $A$ to the object $B$.


If $F$ is full and faithful and injective on objects, we say the $F$ is an embedding. An embedding does not mean that the categories are equal up to isomorphism. With an embedding we are still allowed to have objects in the target category $\mathcal{D}$ that are not present in the source category $\mathcal{C}$. For example let $Cat_1$ be the category with objects $A$ and $B$ and arrows besides the $id$-arrows given as $$ f_1,f_2 : A \rightarrow B $$ Let $Cat_2$ be the category with objects $H$, $I$, $J$ and $K$ and arrows $$ g_1,g_2 : H \rightarrow I $$ along $$ g_3,g_4 : J \rightarrow K $$ Let $F : Cat_1 \rightarrow Cat_2$ the functor that maps objects as follows $$ [A \mapsto H,B \mapsto I] $$ and arrows as $$ [f_1 \mapsto g_1, f_2 \mapsto g_2] $$ Here $F$ is full, faithful and injective on objects. Hence $F$ is an embedding. It makes quite good intuitive sense that $Cat_1$ is embedded in $Cat_2$.

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