quality:correction
Opposite functor

Every functor F: C → D {\displaystyle F\colon C\to D} induces the opposite functor F o p: C o p → D o p {\displaystyle F^{\mathrm {op} }\colon C^{\mathrm {op} }\to D^{\mathrm {op} }}, where C o p {\displaystyle C^{\mathrm {op} }} and D o p {\displaystyle D^{\mathrm {op} }} are the opposite categories to C {\displaystyle C} and D {\displaystyle D}. By definition, F o p {\displaystyle F^{\mathrm {op} }} maps objects and morphisms in the identical way as does F {\displaystyle F}. Since C o p {\displaystyle C^{\mathrm {op} }} does not coincide with C {\displaystyle C} as a category, and similarly for D {\displaystyle D}, F o p {\displaystyle F^{\mathrm {op} }} is distinguished from F {\displaystyle F}. For example, when composing F: C 0 → C 1 {\displaystyle F\colon C_{0}\to C_{1}} with G: C 1 o p → C 2 {\displaystyle G\colon C_{1}^{\mathrm {op} }\to C_{2}}, one should use either G ∘ F o p {\displaystyle G\circ F^{\mathrm {op} }} or G o p ∘ F {\displaystyle G^{\mathrm {op} }\circ F}. Note that, following the property of opposite category, (F o p) o p = F {\displaystyle \left(F^{\mathrm {op} }\right)^{\mathrm {op} }=F}.