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Let C and D be two categories and suppose we can define a contravariant functor between them. I want to show that $F:C \rightarrow D^{op}$ defines a covariant functor, but I can't show $F(g\circ f)=Fg \circ Ff$. How can I show that F preserves composition?

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So you have a contravariant function $G\colon C\to D$, and are defining $F\colon C\to D^{op}$ by $F(x)=G(x)$ for all objects, and $F(f) = G(f)^{op}$ for all arrows, yes? Note that the arrow $g\circ f$ is mapped to the arrow $G(g\circ f)^{op}$, which equals $(G(f)\circ G(g))^{op}$, hence equals $G(g)^{op}\circ G(f)^{op}$, hence equals $F(g)\circ F(f)$ in $D^{op}$. – Arturo Magidin Sep 18 '11 at 22:11
up vote 3 down vote accepted

A contravariant functor $F:C\to D$ sends a morphism $f\in\text{Hom}_C(X,Y)$ to $Ff\in\text{Hom}_D(FY,FX)$, and satisfies $F(g\circ f)=Ff\circ Fg$.

The covariant functor $F^{op}:C\to D^{op}$ is gotten by composing $F$ with the contravariant functor $Op:D\to D^{op}$ that sends a morphism $h\in\text{Hom}_D(P,Q)$ to $Op(h)=h^{op}\in\text{Hom}_{D^{op}}(Q,P)$. Since $Op$ is contravariant, it satisfies $$Op(h\circ k)=(h\circ k)^{op}=k^{op}\circ h^{op}=Op(k)\circ Op(h).$$

Thus $F^{op}=Op\circ F$ satisfies $$F^{op}(g\circ f)=Op(Ff\circ Fg)=(Fg)^{op}\circ (Ff)^{op}=F^{op}(g)\circ F^{op}(f),$$ i.e. $F^{op}$ is covariant.

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