ConstantFunctor 2007-10-24 00:36:33 2008-08-08 19:16:36 Definition functor \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}} Let $\mathcal{C}$ and $\mathcal{D}$ be categories. A \emph{constant functor} from $\mathcal{C}$ to $\mathcal{D}$ is a functor $k:\mathcal{C\to D}$ 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 $1_A\circ 1_A=1_A$. \textbf{Remarks}. \begin{itemize} \item For the constant functor $k$ considered above, the object $A$ is the \emph{fixed value} of $k$. To identify $k$ with $A$, we often write $k_A:=k$, or simply $A$ when no confusion arises. \item Composing a functor with a constant functor gives us a constant functor. More precisely, let $k_A:\mathcal{C\to D}$ be the constant functor with fixed value $A$. If $F:\mathcal{B\to C}$ is a functor, then $k_A\circ F:\mathcal{B\to D}$ is the constant functor $(k\circ F)_A$ with fixed value at $A$. Moreover, if $G:\mathcal{D\to E}$ is a functor, then $G\circ k_A:\mathcal{C\to E}$ is the constant functor $(G\circ k)_{G(A)}$ valued at $G(A)$. \item Given any functor $T:\mathcal{C\to D}$, any natural transformation $\tau_A: k_A\dot{\to} T$ takes any object $X\in \mathcal{C}$ to a morphism $A\to T(X)$, and, for any morphism $\alpha: X\to Y$ in $\mathcal{C}$, a commutative triangle $$\xymatrix@R=7pt@C=1.5cm{ & T(X) \ar[dd]^{T(\alpha)} \\ A \ar[dr] \ar[ur] & \\ & T(Y) } $$ in $\mathcal{D}$. Similarly, any natural transformation $\sigma_A:T\dot{\to} k_A$ takes any object $X$ to $T(X)\to A$ and any morphism $\alpha: X\to Y$ to a commutative triangle $$\xymatrix@R=7pt@C=1.5cm{ T(X) \ar[dd]_{T(\alpha)} \ar[dr] & \\ & A \\ T(Y) \ar[ur] & } $$ \item If $\alpha:A\to B$ is any morphism in $\mathcal{D}$, then the natural transformation $\tau:k_A\dot{\to} k_B$ given by sending every object $X\in \mathcal{C}$ to the morphism $\tau_X: A\to B:=\alpha$ can be thought of as a ``constant'' natural transformation, since, for any morphism $\beta:X\to Y$ in $\mathcal{C}$, the commutative diagram $$ \xymatrix@R=1.25cm@C=2cm{ k_A(X) \ar[d]_{k_A\beta} \ar[r]^{\tau_X} & \ar@{}[dr]|{=} k_B(X) \ar[d]^{k_B\beta} & A \ar[d]_{1_A} \ar[r]^{\alpha} & B \ar[d]^{1_B} \\ k_A(Y) \ar[r]^{\tau_Y} & k_B(Y) & A \ar[r]^{\alpha} & B } $$ reduces to $\alpha$. \item As above, denote $\tau$ by $\tau_{\alpha}$. Let $\sigma_A: T\dot{\to} k_A$ be a natural transformation. Then the transformation $\tau_{\alpha}\circ \sigma_A:T\dot{\to} k_B$ sends any object $X$ in $\mathcal{C}$ to the morphism $$T(X)\to A \stackrel{\alpha}{\to} B$$ in $\mathcal{D}$, and any morphism $\beta:X\to Y$ to the commutative diagram $$\xymatrix@R=7pt@C=1.5cm{ T(X) \ar[dd]_{T(\beta)} \ar[dr] & & \\ & A \ar[r]^{\alpha} & B \\ T(Y) \ar[ur] & & } $$ For any natural transformation $\gamma_B: k_B\dot{\to} T$, the composition $\sigma_B\circ \tau_{\alpha}$ works similarly. \end{itemize} \begin{thebibliography}{9} \bibitem{Ma}S. Mac Lane, \emph{Categories for the Working Mathematician} (2nd edition), Springer-Verlag, 1997. \end{thebibliography}