A constant functor$\Delta d: C\to D$ is a functor that maps each object of the category$C$ to a fixed object $d\in D$ and each morphism of $C$ to the identity morphism of that fixed object.

(The notation $\Delta d$ is suggested by the fact that if $d: 1 \to D$ names an object $d$ of $D$, then the composite

$1 \stackrel{d}{\to} D \stackrel{\Delta}{\to} D^C$

names a functor $\Delta d: C \to D$ which is the constant functor at $d$. Here $\Delta: D \to D^C$ denotes a diagonal functor.)

Viewing $D \cong [1,D]$, it is possible to see that $\Delta$ is a base change of the functor categories $D^1$ and $D^C$, given the functor $!_C: C \to 1$.

More generally, we may say that a functor $F:C\to D$ is essentially constant if it is naturally isomorphic to a constant functor.

Properties

Note that a constant functor can be expressed as the composite

$C \stackrel{!}{\to} 1 \stackrel{[d]}{\to} D.$

Here $1$ is a terminal category (exactly one object and exactly one morphism, namely the identity), and $[d]$ denotes the unique functor from $1$ with $F(\bullet) = d$ and $F(Id_\bullet) = Id_d$. It follows that $F:C\to D$ is essentially constant if and only if it factors through $1$ up to isomorphism.

If $C$ has at least one object, then there is a codescent object

$C\times C\times C \underoverset{\to}{\to}{\to} C\times C \rightrightarrows C \to 1.$

Therefore, in this case a functor $F:C\to D$ is essentially constant if and only if we have a natural isomorphism between the two composites $C\times C \rightrightarrows C \xrightarrow{F} D$ (i.e. isomorphisms $F(c_1)\cong F(c_2)$ natural in $c_1,c_2\in C$) that satisfies a cocycle condition?. This is a categorified version of the statement that a function with inhabited domain is a constant function if and only if the images of any two elements are equal. Just as in that case, there is a distinction in the trivial case: the identity functor of the empty category satisfies this weaker condition, but is not essentially constant because there is no object for it to be constant at.