The diagonal functor is a categorification of the diagonal function.

Let $C$ be a category. The **(binary) diagonal functor of $C$** is the functor $\Delta\colon C \to C \times C$ given by $\Delta(x) = (x,x)$, regardless of whether $x$ is an object or an arrow of $C$.

More generally, let $J$ and $C$ be arbitrary categories. The **$J$-ary diagonal functor of $C$** is the functor $\Delta_J\colon C\to C^J$ sending each object $c$ to the constant functor $\Delta c$ (the functor having value $c$ for each object of $J$ and value $1_c$ for each arrow of $J$), and each arrow $f\colon c\to c'$ of $C$ to the natural transformation $\Delta f\colon \Delta c \stackrel{.}{\to} \Delta c'$ which has the same value $f$ at each object $j$ of $J$.

Since $C$ is $J$-cocomplete ($J$-complete) iff $\Delta$ has a left (right) adjoint, the general adjoint functor theorem may be used in some cases to prove cocompleteness (completeness). For this to work, $\Delta$ must at least preserve small limits (colimits).

Let $P$ and $C$ be arbitrary categories. Then $\Delta_P\colon C\to C^P$ preserves *all* limits that exist in $C$.

First, recall that limits in functor categories are calculated pointwise. In some detail, if for an object $p\in \mathrm{obj}(P)$ we write $E_p:X^P\to X$ for the ‘’evaluate at $p$’‘ functor (with $E_p(H\colon P\to X)=H(p)$ and $E_p(\sigma\colon H\stackrel{.}{\to} H')=\sigma_p\colon H(p)\to H'(p)$), then we have the following fact (Theorem V.3.1 on p. 115 of Categories Work): If $S\colon J\to X^P$ is such that for each object $p$ of $P$, $E_p S\colon J\to X$ has a limiting cone $\tau_p\colon L(p)\stackrel{.}{\to} E_p S$, then there exists a unique functor $L$ with object function $p\mapsto L(p)$ such that $\tilde{\tau}=\{\tilde{\tau}_{j,p}\}$ with $\tilde{\tau}_{j,p}:=\tau_{p,j}$ is a cone $\tilde{\tau}\colon \Delta_J(L)\stackrel{.}{\to} S$; moreover, this $\tilde{\tau}$ is a limiting cone from $L\in \mathrm{obj}(X^P)$ to $S\colon J\to X^P$.

Back to the proof of the proposition, let $F\colon J\to C$ be a functor with a limiting cone $\nu\colon \Delta_J(\ell) \stackrel{.}{\to} F$. We would like to show that $\Delta_P\nu\colon \Delta_P\circ \bigl(\Delta_J(\ell)\bigr) \stackrel{.}{\to} \Delta_P\circ F$ is a limiting cone. Noting that $\Delta_P\circ \bigl(\Delta_J(\ell)\bigr)=\Delta_J(\Delta_P(\ell))$ (where the first $\Delta_J$ is $C\to C^J$ and the second is $C^P\to (C^P)^J$), the last cone may be written as $\Delta_P\nu\colon \Delta_J(\Delta_P(\ell)) \stackrel{.}{\to} \Delta_P\circ F$.

First, we note that for each object $p$ of $P$, $E_p\circ(\Delta_P\circ F)$ is just $F$, and therefore has the limiting cone $\nu\colon\ell \stackrel{.}{\to} F$ by assumption. Hence, it is clear that $\Delta_P\circ F$ has a limit, but we must verify that $\Delta_P\nu$ is a limiting cone.

One functor $P\to X$ with object function $p\mapsto \ell$ is just $\Delta_P(\ell)$. For this functor, we have our cone $\Delta_P\nu\colon \Delta_J(\Delta_P(\ell)) \stackrel{.}{\to} \Delta_P\circ F$. Since for all $j$ and $p$ we have $(\Delta_P\nu)_{j,p}=\nu_j=j\text{th component of the limiting cone of }E_p\circ(\Delta_P\circ F)$, we are done by the theorem on pointwise limits.

Last revised on August 20, 2018 at 12:51:28. See the history of this page for a list of all contributions to it.