Let $\mathsf{C}$ be any category. The opposite category $\mathsf{C}^\mathrm{op}$ has\n1. the same objects as in $\mathsf{C}$, and\n2. a morphism $f^\mathrm{op}$ in $\mathsf{C}^\mathrm{op}$ for each a morphism $f$ in $\mathsf{C}$ so that the domain of $f^\mathrm{op}$ is defined to be the codomain of $f$ and the codomain of $f^\mathrm{op}$ is defined to be the domain of $f$: i.e.,\n\(f^{\mathrm{op}} \colon X \to Y\quad \in \mathsf{C}^\mathrm{op} \qquad \leftrightsquigarrow \qquad f \colon Y \to X\quad \in \mathsf{C}.\)\nThat is, $\mathsf{C}^\mathrm{op}$ has the same objects and morphisms as $\mathsf{C}$, except that ``each morphism is pointing in the opposite direction.’’ The remaining structure of the category $\mathsf{C}^\mathrm{op}$ is given as follows:\n1. For each object $X$, the arrow $1_X^\mathrm{op}$ serves as its identity in $\mathsf{C}^\mathrm{op}$.\n2. To define composition, observe that a pair of morphisms $f^\mathrm{op},g^\mathrm{op}$ in $\mathsf{C}^\mathrm{op}$ is composable precisely when the pair $g,f$ is composable in $\mathsf{C}$, i.e., precisely when the codomain of $g$ equals the domain of $f$. We then define $g^\mathrm{op} \cdot f^\mathrm{op}$ to be $(f\cdot g)^\mathrm{op}$: i.e.,\n\(\begin{array}{clccl}\n f^\mathrm{op} \colon X \to Y,\ g^\mathrm{op} \colon Y \to Z & \in \mathsf{C}^\mathrm{op}&\rightsquigarrow& g^\mathrm{op} f^\mathrm{op} \colon X \to Z & \in \mathsf{C}^\mathrm{op}\\ \rotatebox{90}{$\leftrightsquigarrow$} & & & \rotatebox{90}{$\leftrightsquigarrow$}& \\ g \colon Z \to Y,\ f \colon Y \to X &\in \mathsf{C} & \rightsquigarrow & f g \colon Z \to X & \in \mathsf{C}\n \end{array}\)