Composite Functor is Functor

Definition

Let $\mathbf C, \mathbf D$ and $\mathbf E$ be metacategories.

Let $F: \mathbf C \to \mathbf D$ and $G: \mathbf D \to \mathbf E$ be covariant functors.

Let $GF: \mathbf C \to \mathbf E$ be the composition of $G$ with $F$.

Then $G F$ is also a covariant functor.

Proof

Let $f, g$ be morphisms of $\mathbf C$ such that $g \circ f$ is defined.

Then:

\(\ds \map {G F} {g \circ f}\)

\(=\)

\(\ds \map G {\map F {g \circ f} }\)

Definition of Composition of Functors

\(\ds \)

\(=\)

\(\ds \map G {F g \circ F f}\)

$F$ is a Covariant Functor

\(\ds \)

\(=\)

\(\ds \map G {F g} \circ \map G {F f}\)

$G$ is a functor

\(\ds \)

\(=\)

\(\ds G F g \circ G F f\)

Definition of Composition of Functors

Also, for any object $C$ of $\mathbf C$:

\(\ds G F I_C\)

\(=\)

\(\ds \map G {F I_C}\)

Definition of Composition of Functors

\(\ds \)

\(=\)

\(\ds G I_{F C}\)

$F$ is a Covariant Functor

\(\ds \)

\(=\)

\(\ds I_{\map G {F C} }\)

$G$ is a Covariant Functor

\(\ds \)

\(=\)

\(\ds I_{G F C}\)

Definition of Composition of Functors

Hence $G F$ is shown to be a covariant functor.

$\blacksquare$