A description/proof of that categories equivalence is equivalence relation
Topics
About: category
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that equivalence of categories is an equivalence relation.
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.
Main Body
1: Description
Equivalence of categories is an equivalence relation.
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, \(F_{1, 1}: C_1 \rightarrow C_1\) and \(F_{1, 1}: C_1 \rightarrow C_1\), 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, \(F_{1, 2}: C_1 \rightarrow C_2\) and \(F_{2, 1}: C_2 \rightarrow C_1\), 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, \(F_{2, 1}: C_2 \rightarrow C_1\) and \(F_{1, 2}: C_1 \rightarrow C_2\), 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 \(F_{1, 2}: C_1 \rightarrow C_2\) and \(F_{2, 1}: C_2 \rightarrow C_1\) 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 \(F_{2, 3}: C_2 \rightarrow C_3\) and \(F_{3, 2}: C_3 \rightarrow C_2\) 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 \(F_{1, 3} = F_{2, 3} F_{1, 2}: C_1 \rightarrow C_2 \rightarrow C_3\) and \(F_{3, 1} = F_{2, 1} F_{3, 2}: C_3 \rightarrow C_2 \rightarrow C_1\).
Is \(F_{3, 1} F_{1, 3} \cong 1_{C_1}\)? Are there some natural transformations, \(\eta_{1, 3}: F_{3, 1} F_{1, 3} \rightarrow 1_{C_1}\) and \(\theta_{1, 3}: 1_{C_1} \rightarrow F_{3, 1} F_{1, 3}\), 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 \(\eta_{2, 3, F_{1, 2} (c)}: F_{3, 2} F_{2, 3} (F_{1, 2} (c)) \rightarrow F_{1, 2} (c)\), \(F_{2, 1} (\eta_{2, 3, F_{1, 2} (c)}): F_{2, 1} F_{3, 2} F_{2, 3} (F_{1, 2} (c)) \rightarrow F_{2, 1} F_{1, 2} (c)\), and \(\eta_{1, 2, c}: F_{2, 1} F_{1, 2} (c) \rightarrow c\) while \(\eta_{1, 3, c}: F_{3, 1} F_{1, 3} (c) = F_{2, 1} F_{3, 2} F_{2, 3} F_{1, 2} (c) \rightarrow c\). Is \(\eta_{1, 3}\) really a natural transformation? For any morphism in \(C_1\), \(\alpha: c \rightarrow c'\), \(\alpha \eta_{1, 3, c} = \eta_{1, 3, c'} F_{3, 1} F_{1, 3} (\alpha) = \eta_{1, 3, c'} F_{2, 1} F_{3, 2} F_{2, 3} F_{1, 2} (\alpha)\)? The left hand side is \(\alpha \eta_{1, 2, c} F_{2, 1} (\eta_{2, 3, F_{1, 2} (c)}) = \eta_{1, 2, c'} F_{2, 1} F_{1, 2} (\alpha) F_{2, 1} (\eta_{2, 3, F_{1, 2} (c)}) = \eta_{1, 2, c'} F_{2, 1} (F_{1, 2} (\alpha) \eta_{2, 3, F_{1, 2} (c)})\). The right hand side is \(\eta_{1, 2, c'} F_{2, 1} (\eta_{2, 3, F_{1, 2} (c')}) F_{2, 1} F_{3, 2} F_{2, 3} F_{1, 2} (\alpha) = \eta_{1, 2, c'} F_{2, 1} (\eta_{2, 3, F_{1, 2} (c')} F_{3, 2} F_{2, 3} F_{1, 2} (\alpha))\). But \(F_{1, 2} (\alpha) \eta_{2, 3, F_{1, 2} (c)} = \eta_{2, 3, F_{1, 2} (c')} F_{3, 2} F_{2, 3} F_{1, 2} (\alpha)\), 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 \(\alpha \eta_{1, 3, c} = \eta_{1, 3, c'} F_{3, 1} F_{1, 3} (\alpha)\), \(\theta_{1, 3, c'} \alpha \eta_{1, 3, c} \theta_{1, 3, c} = \theta_{1, 3, c'} \eta_{1, 3, c'} F_{3, 1} F_{1, 3} (\alpha) \theta_{1, 3, c}\), \(\theta_{1, 3, c'} \alpha = F_{3, 1} F_{1, 3} (\alpha) \theta_{1, 3, c}\).
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.
