---- > [!definition] Definition. ([[opposite category]]) > Let $\mathsf{C}$ be a [[category]]. The **opposite category** $\mathsf{C}^{\text{op}}$ is defined so that >- $\text{Obj}(\mathsf{C}^{\text{op}}):= \text{Obj}(\mathsf{C})$; >- for $A,B$ objects of $\mathsf{C}^{\text{op}}$, $\text{Hom}_{\mathsf{C}^{\text{op}}}(A,B):=\text{Hom}_{\mathsf{C}}(B,A)$, > with composition given by 'reversing arrows' in $\mathsf{C}$: the morphisms $f \in \text{Hom}_{\mathsf{C}^{\text{op}}}(A,B)$, $g \in \text{Hom}_{\mathsf{C}^{\text{op}}}(B,C)$ determine the morphism $gf \in \text{Hom}_{\mathsf{C}}(C,A)=\text{Hom}_{\mathsf{C}^{\text{op}}}(A,C)$. ^definition > [!justification] > Given an object $A$ in $\mathsf{C}$, we have $\text{Hom}_{\mathsf{C}^{\text{op}}}(A,A)=\text{End}_{\mathsf{C}^{\text{op}}}(A)=\text{Hom}_{\mathsf{C}}(A,A)$, and so the identity morphism $1_{A} \in \text{End}_{\mathsf{C}^{\text{op}}}(A)$ is unchanged from $\mathsf{C}$. Given $f \in \text{Hom}_{\mathsf{C}^{\text{op}}}(A,B)$, we have that $f=f'$ for some $f' \in \text{Hom}_{\mathsf{C}}(B,A)$ and so $f 1_{A}=1_{A}f'=f'=f$. Likewise for left identities and so identities and composition interact correctly. > To verify associativity, let $f \in \text{Hom}_{\mathsf{C}^{\text{op}}}(A,B)$, $g \in \text{Hom}_{\mathsf{C}^{\text{op}}}(B,C)$, $h \in \text{Hom}_{\mathsf{C}^{\text{op}}}(C,D)$ and denote by $f'=f$, $g'=g$, $h'=h$ the counterparts from $\mathsf{C}$. We have $f(gh)=(h'g')f'=h'(g'f')=(fg)h \ (?)$ as required. ^justification ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` #reformatrevisebatch02