[proofplan] We verify the functor axioms directly. First we check that the composite assignment sends each morphism of $\mathcal C$ to a morphism of $\mathcal E$ with the correct source and target. Then we prove that identities and composition are preserved by applying the corresponding axioms first for $F$ and then for $G$. Finally, we verify the same two axioms for the identity assignment on an arbitrary category. [/proofplan] [step:Check that the composite sends morphisms to morphisms with the correct source and target] Let $H: \mathcal C \to \mathcal E$ denote the assignment $G \circ F$. Let $X$ and $Y$ be objects of $\mathcal C$, and let $f: X \to Y$ be a morphism in $\mathcal C$. Since $F: \mathcal C \to \mathcal D$ is a covariant functor, $F(f): F(X) \to F(Y)$ is a morphism in $\mathcal D$. Since $G: \mathcal D \to \mathcal E$ is a covariant functor, applying $G$ to this morphism gives \begin{align*} G(F(f)): G(F(X)) \to G(F(Y)) \end{align*} as a morphism in $\mathcal E$. By the definition of $H$, this says \begin{align*} H(f): H(X) \to H(Y). \end{align*} Thus $H$ is well-defined on morphisms with the required source and target. [/step] [step:Prove that the composite preserves identity morphisms] Let $X$ be an object of $\mathcal C$. Since $F$ is a covariant functor, it preserves identity morphisms, so \begin{align*} F(\operatorname{id}_X)=\operatorname{id}_{F(X)}. \end{align*} Since $G$ is a covariant functor, it also preserves identity morphisms, so \begin{align*} G(\operatorname{id}_{F(X)})=\operatorname{id}_{G(F(X))}. \end{align*} Using the definition of $H=G \circ F$, we obtain \begin{align*} H(\operatorname{id}_X) &=G(F(\operatorname{id}_X)) \\ &=G(\operatorname{id}_{F(X)}) \\ &=\operatorname{id}_{G(F(X))} \\ &=\operatorname{id}_{H(X)}. \end{align*} Therefore $H$ preserves identity morphisms. [/step] [step:Prove that the composite preserves composition of morphisms] Let $X$, $Y$, and $Z$ be objects of $\mathcal C$. Let $f: X \to Y$ and $g: Y \to Z$ be morphisms in $\mathcal C$, so that $g \circ f: X \to Z$ is defined in $\mathcal C$. Since $F$ is a covariant functor, it preserves composition: \begin{align*} F(g \circ f)=F(g)\circ F(f). \end{align*} The morphisms $F(f): F(X) \to F(Y)$ and $F(g): F(Y) \to F(Z)$ are composable in $\mathcal D$. Since $G$ is a covariant functor, it preserves their composition: \begin{align*} G(F(g)\circ F(f))=G(F(g))\circ G(F(f)). \end{align*} Using the definition of $H=G \circ F$, we get \begin{align*} H(g \circ f) &=G(F(g \circ f)) \\ &=G(F(g)\circ F(f)) \\ &=G(F(g))\circ G(F(f)) \\ &=H(g)\circ H(f). \end{align*} Thus $H$ preserves composition. [/step] [step:Conclude that the composite assignment is a covariant functor] The assignment $H=G \circ F$ is defined on objects and morphisms, sends every morphism $f: X \to Y$ in $\mathcal C$ to a morphism $H(f): H(X) \to H(Y)$ in $\mathcal E$, preserves identity morphisms, and preserves composition of composable morphisms. These are precisely the defining properties of a covariant functor. Hence $G \circ F: \mathcal C \to \mathcal E$ is a covariant functor. [/step] [step:Verify that the identity assignment is a covariant functor] Let $\mathcal C$ be a category. Define $I: \mathcal C \to \mathcal C$ by $I(X)=X$ for every object $X$ of $\mathcal C$ and $I(f)=f$ for every morphism $f$ of $\mathcal C$. If $f: X \to Y$ is a morphism in $\mathcal C$, then \begin{align*} I(f)=f: X \to Y, \end{align*} so $I(f): I(X) \to I(Y)$ has the correct source and target. For every object $X$ of $\mathcal C$, \begin{align*} I(\operatorname{id}_X)=\operatorname{id}_X=\operatorname{id}_{I(X)}. \end{align*} For every pair of composable morphisms $f: X \to Y$ and $g: Y \to Z$ in $\mathcal C$, \begin{align*} I(g \circ f)=g \circ f=I(g)\circ I(f). \end{align*} Therefore $I$ preserves identity morphisms and composition. Hence $I=\operatorname{Id}_{\mathcal C}$ is a covariant functor. [/step]