|
|
|
Viewing Version
1
of
'constant functor'
|
[ view 'constant functor'
|
back to history
]
| Title of object: |
constant functor |
| Canonical Name: |
ConstantFunctor |
| Type: |
Definition |
| Created on: |
2007-10-24 00:36:33 |
| Modified on: |
2007-10-24 00:36:33 |
| Classification: |
msc:18-00 |
Preamble:
\usepackage{amssymb,amscd}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{mathrsfs}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
\usepackage{amsthm}
% making logically defined graphics
\usepackage{xypic}
\usepackage{pst-plot}
\usepackage{psfrag}
% define commands here
\newtheorem{prop}{Proposition}
\newtheorem{thm}{Theorem}
\newtheorem{ex}{Example}
\newcommand{\real}{\mathbb{R}}
\newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}} |
Content:
Let $\mathcal{C}$ and $\mathcal{D}$ be categories. A \emph{constant functor} from $\mathcal{C}$ to $\mathcal{D}$ is a functor $k:\mathcal{C}\to \mathcal{D}$ such that there is an object $A\in \mathcal{D}$ such that
\begin{itemize}
\item for all objects $X$ in $\mathcal{C}$, $k(X)=A$, and
\item for all morphisms $X\to Y$ in $\mathcal{C}$, $k(X\to Y)=1_A$, the identity morphism of $A$.
\end{itemize}
To see that this is indeed a functor, we merely need to verify that $$k\big((X\to Y)\circ (Y\to Z)\big)=k(X\to Y)\circ k(Y\to Z).$$ But this is obvious, as the left hand side is $k(X\to Z)=1_A$, while the right hand side is $1_A\circ 1_A=1_A$. |
|
|
|
|
|