Definition. Given a category $\mathcal{C}$, the opposite category $\mathcal{C}^{\text{op}}$ is defined in the following way.
- $\text{Ob}(\mathcal{C}^{\text{op}}):=\text{Ob}(\mathcal{C})$.
- For $X,Y\in\mathcal{C}$, $\mathcal{C}^{\text{op}}(X,Y):=\mathcal{C}(Y,X)$ (to avoid confusion, we shall write $f^{\text{op}}:X\rightarrow Y$ for the morphism of $\mathcal{C}^{\text{op}}$ corresponding to the morphism $f:Y\rightarrow X$ of $\mathcal{C}$).
- The composition law of $\mathcal{C}^{\text{op}}$ is given by $$(f^{\text{op}}\circ g^{\text{op}}):=(g\circ f)^{\text{op}}$$ for morphisms $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ of $\mathcal{C}$.
Let $F:\mathcal{C}\rightarrow\mathcal{D}$ be a contravariant functor. I want to show that $F$ is a covariant functor from $\mathcal{C}^{\text{op}}$ to $\mathcal{D}$. W.r.t. the mapping of objects nothing changes. For $X,Y\in\mathcal{C}^{\text{op}}$, we have a mapping $$\mathcal{C}^{\text{op}}(X,Y)=\mathcal{C}(Y,X)\rightarrow\mathcal{D}(F(X),F(Y)),\,f\mapsto F(f).$$ Let $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ be arrows of $\mathcal{C}^{\text{op}}$. We have to show that $F(g\circ f)=F(g)\circ F(f)$. Since $f\in\text{C}^{\text{op}}(X,Y)$ and $g\in\mathcal{C}^{\text{op}}(Y,Z)$, it is the case that $f^{\text{op}}\in\mathcal{C}(Y,X)$ and $g^{\text{op}}\in\mathcal{C}(Z,Y)$. Then $$F(g\circ f)=F((g^{\text{op}})^{\text{op}}\circ(f^{\text{op}})^{\text{op}})=F((f^{\text{op}}\circ g^{\text{op}})^{\text{op}})=?$$ I'm now stuck. Can someone please help complete this using Borceux's definition.