381: Categories Equivalence Is Equivalence Relation

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A description/proof of that categories equivalence is equivalence relation

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Topics

About: category

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of equivalence of categories.
• The reader knows a definition of equivalence relation.
• The reader admits the proposition that any equivalence functor preserves and reflects monics, epics, bimorphisms, split monics, split epics, isomorphisms, and commutative diagrams.

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Target Context

• The reader will have a description and a proof of the proposition that equivalence of categories is an equivalence relation.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

Equivalence of categories is an equivalence relation.

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2: Proof

Let $C_1,C_2,C_3$ be any categories such that $C_1$ is equivalent with $C_2$ and $C_2$ is equivalent with $C_3$.

Let us check the reflexivity. Is $C_1$ equivalent with $C_1$? Let us take the functors, $F_{1,1}:C_1\to C_1$ and $F_{1,1}:C_1\to C_1$, which means that same $F_{1,1}$ is for the both directions, that are the identity functor, $1_{C_1}$. Is there a natural isomorphism between $F_{1,1}{F}_{1,1}$ and $1_{C_1}$? $F_{1,1}{F}_{1,1}=1_{C_1}$. For the natural transformations, $\eta :1_{C_1}\to 1_{C_1}$ and $\theta :1_{C_1}\to 1_{C_1}$, let us take the identity natural transformation, $1_{1_{C_1}}$. $\theta \eta =1_{1_{C_1}}{1}_{1_{C_1}}=1_{1_{C_1}}=1_{1_{C_1}}{1}_{1_{C_1}}=\eta \theta$. So, $\eta$ is a natural isomorphism.

Let us check the symmetry. Is $C_2$ equivalent with $C_1$? There are some functors, $F_{1,2}:C_1\to C_2$ and $F_{2,1}:C_2\to C_1$, such that $F_{2,1}{F}_{1,2}\cong 1_{C_1}$ and $F_{1,2}{F}_{2,1}\cong 1_{C_2}$. So, there are the functors, $F_{2,1}:C_2\to C_1$ and $F_{1,2}:C_1\to C_2$, such that $F_{1,2}{F}_{2,1}\cong 1_{C_2}$ and $F_{2,1}{F}_{1,2}\cong 1_{C_1}$.

Let us check the transitivity. Is $C_1$ equivalent with $C_3$? There are $F_{1,2}:C_1\to C_2$ and $F_{2,1}:C_2\to C_1$ such that $F_{2,1}{F}_{1,2}\cong 1_{C_1}$ with some natural isomorphisms, ${\eta }_{1,2}:F_{2,1}{F}_{1,2}\to 1_{C_1},{\theta }_{1,2}:1_{C_1}\to F_{2,1}{F}_{1,2}$, and $F_{1,2}{F}_{2,1}\cong 1_{C_2}$ with some natural isomorphisms, ${\eta }_{2,1}:F_{1,2}{F}_{2,1}\to 1_{C_2},{\theta }_{2,1}:1_{C_2}\to F_{1,2}{F}_{2,1}$. There are $F_{2,3}:C_2\to C_3$ and $F_{3,2}:C_3\to C_2$ such that $F_{3,2}{F}_{2,3}\cong 1_{C_2}$ with some natural isomorphisms, ${\eta }_{2,3}:F_{3,2}{F}_{2,3}\to 1_{C_2},{\theta }_{2,3}:1_{C_2}\to F_{3,2}{F}_{2,3}$, and $F_{2,3}{F}_{3,2}\cong 1_{C_3}$ with some natural isomorphisms, ${\eta }_{3,2}:F_{2,3}{F}_{3,2}\to 1_{C_3},{\theta }_{3,2}:1_{C_3}\to F_{2,3}{F}_{3,2}$.

Let us take $F_{1,3}=F_{2,3}{F}_{1,2}:C_1\to C_2\to C_3$ and $F_{3,1}=F_{2,1}{F}_{3,2}:C_3\to C_2\to C_1$.

Is $F_{3,1}{F}_{1,3}\cong 1_{C_1}$? Are there some natural transformations, ${\eta }_{1,3}:F_{3,1}{F}_{1,3}\to 1_{C_1}$ and ${\theta }_{1,3}:1_{C_1}\to F_{3,1}{F}_{1,3}$, such that ${\theta }_{1,3}{\eta }_{1,3}=1_{F_{3,1}{F}_{1,3}}$ and ${\eta }_{1,3}{\theta }_{1,3}=1_{1_{C_1}}$? Let us define ${\eta }_{1,3,c}:={\eta }_{1,2,c}{F}_{2,1}({\eta }_{2,3,F_{1,2}(c)})$, which makes sense because ${\eta }_{2,3,F_{1,2}(c)}:F_{3,2}{F}_{2,3}(F_{1,2}(c))\to F_{1,2}(c)$, $F_{2,1}({\eta }_{2,3,F_{1,2}(c)}):F_{2,1}{F}_{3,2}{F}_{2,3}(F_{1,2}(c))\to F_{2,1}{F}_{1,2}(c)$, and ${\eta }_{1,2,c}:F_{2,1}{F}_{1,2}(c)\to c$ while ${\eta }_{1,3,c}:F_{3,1}{F}_{1,3}(c)=F_{2,1}{F}_{3,2}{F}_{2,3}{F}_{1,2}(c)\to c$. Is ${\eta }_{1,3}$ really a natural transformation? For any morphism in $C_1$, $\alpha :c\to c^{\prime }$, $\alpha {\eta }_{1,3,c}={\eta }_{1,3,c^{\prime }}{F}_{3,1}{F}_{1,3}(\alpha )={\eta }_{1,3,c^{\prime }}{F}_{2,1}{F}_{3,2}{F}_{2,3}{F}_{1,2}(\alpha )$? The left hand side is $\alpha {\eta }_{1,2,c}{F}_{2,1}({\eta }_{2,3,F_{1,2}(c)})={\eta }_{1,2,c^{\prime }}{F}_{2,1}{F}_{1,2}(\alpha )F_{2,1}({\eta }_{2,3,F_{1,2}(c)})={\eta }_{1,2,c^{\prime }}{F}_{2,1}(F_{1,2}(\alpha ){\eta }_{2,3,F_{1,2}(c)})$. The right hand side is ${\eta }_{1,2,c^{\prime }}{F}_{2,1}({\eta }_{2,3,F_{1,2}(c^{\prime })})F_{2,1}{F}_{3,2}{F}_{2,3}{F}_{1,2}(\alpha )={\eta }_{1,2,c^{\prime }}{F}_{2,1}({\eta }_{2,3,F_{1,2}(c^{\prime })}{F}_{3,2}{F}_{2,3}{F}_{1,2}(\alpha ))$. But $F_{1,2}(\alpha ){\eta }_{2,3,F_{1,2}(c)}={\eta }_{2,3,F_{1,2}(c^{\prime })}{F}_{3,2}{F}_{2,3}{F}_{1,2}(\alpha )$, so, the both sides equal, so, ${\eta }_{1,3}$ is a natural transformation. ${\eta }_{1,3,c}$ is an isomorphism, because ${\eta }_{2,3,F_{1,2}(c)}$ is so, the equivalence functor, $F_{2,1}$, preserves isomorphism, by the proposition that any equivalence functor preserves and reflects monics, epics, bimorphisms, split monics, split epics, isomorphisms, and commutative diagrams, and ${\eta }_{1,2,c}$ is so. Then, there is the inverse of ${\eta }_{1,3,c}$, ${{\eta }_{1,3,c}}^{-1}$, and ${\theta }_{1,3,c}={{\eta }_{1,3,c}}^{-1}$. ${\theta }_{1,3}$ is a natural transformation, because from $\alpha {\eta }_{1,3,c}={\eta }_{1,3,c^{\prime }}{F}_{3,1}{F}_{1,3}(\alpha )$, ${\theta }_{1,3,c^{\prime }}\alpha {\eta }_{1,3,c}{\theta }_{1,3,c}={\theta }_{1,3,c^{\prime }}{\eta }_{1,3,c^{\prime }}{F}_{3,1}{F}_{1,3}(\alpha ){\theta }_{1,3,c}$, ${\theta }_{1,3,c^{\prime }}\alpha =F_{3,1}{F}_{1,3}(\alpha ){\theta }_{1,3,c}$.

Is $F_{1,3}{F}_{3,1}\cong 1_{C_3}$? Yes: by the symmetry, $C_3$ is equivalent with $C_2$ and $C_2$ is equivalent with $C_1$, and the symmetric argument of the previous paragraph can be used here.

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References

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