Definition 1.2.7 (Functor).label Let $\frak{C},\frak{D}$ be categories. A covariant (resp. contravariant) functor is a rule which

  • associates each object $X\in \ob{\frak{C}}$ with an object $F(X)\in \ob{\frak{D}}$

  • associates each morphism $f\in \mor{\frak{C}}{X}{Y}$ with a morphism $F(f)\in \mor{\frak{D}}{F(X)}{F(Y)}$ (resp. $\mor{\frak{D}}{F(Y)}{F(X)}$) such that:

    1. (F1)

      $F(\text{Id}_{X})=\text{Id}_{F(X)}$ for each $X\in \ob{\frak{X}}$

    2. (F2)

      $F(g\circ f)=F(g)\circ F(f)$ (resp. $=F(f)\circ F(g)$) for each morphisms $f\in \mor{\frak{C}}{X}{Y},g\in \mor{\frak{C}}{Y}{Z}$

We denote the collection of functors $\text{Fun}_{\text{cov}}\parens{\frak{C},\frak{D}}$ (resp. $\text{Fun}_{\text{contra}}\parens{\frak{C},\frak{D}}$). Then there is a canonical bijection $\text{Fun}_{\text{contra}}\parens{\frak{C},\frak{D}}\to \text{Fun}_{\text{cov}}\parens{\frak{C}^{\text{op}},\frak{D}}$.

We say that $F$ is an endofunctor if $\frak{C}=\frak{D}$. In which case we say that $F$ is involutive if $F\circ F$ is naturally isomorphic to $\indi{\frak{C}}$ and strictly involutive if $F\circ F=\indi{\frak{C}}$.

Proof. Consider the map

\begin{align*}\Phi_{\frak{C},\frak{D}}:\text{Fun}_{\text{contra}}\parens{\frak{C},\frak{D}}\to \text{Fun}_{\text{cov}}\parens{\frak{C}^{\text{op}},\frak{D}}\end{align*}

to be continued$\square$

Post a Comment

Name:Email:
Please enter the tag of the current page (4M) to post the comment.
Tag: