The Clowder Project

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  Item 1a$\implies $Item 1b.

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  Item 1b$\implies $Item 1a.

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  Item 1a$\implies $Item 1c.

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  Item 1c$\implies $Item 1a.

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  Item 1a$\iff $Item 1d$\iff $Item 1e.

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  Faithfulness: Let $f,g\in \operatorname {\mathrm{Hom}}_{\mathcal{C}}(A,B)$ with $F(f)=F(g)$. Then $GF(f)=GF(g)$ and we have the following diagrams:

  
  Since $\eta ^{-1}$ is natural, these commute. We then have
  \begin{align*} f\circ \eta ^{-1}_{A} & = \eta ^{-1}_{B}\circ GF(f)\\ & = \eta ^{-1}_{B}\circ GF(g)\\ & = g\circ \eta ^{-1}_{A}. \end{align*}

  Precomposing both sides with $\eta _{A}$ then gives $f=g$, showing $F$ to be faithful.

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  Fullness: Given $f\in \operatorname {\mathrm{Hom}}_{\mathcal{D}}(F(A),F(B))$, define a morphism $h\colon A\to B$ by

  \[ h\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\eta ^{-1}_{B}\circ G(f)\circ \eta _{A}, \]

  yielding the commutative diagram

  
  We must show $F(h) = f$. Since $(G,F,\epsilon ^{-1},\eta ^{-1})$ is also an equivalence due to Item 3, the functor $G$ is faithful, so it suffices to show $GF(h) = G(f)$.

  From the naturality of $\eta $ applied to $\eta _{A}\colon A\to GF(A)$, we have

  
  for all $A\in \operatorname {\mathrm{Obj}}(\mathcal{C})$. Since $\eta $ is an isomorphism, this implies $\eta _{GF(A)} = GF(\eta _{A})$.

  By applying $GF$ to the diagram $(\dagger )$ and incorporating the equality, we see that the diagram

  
  must commute. By naturality of $\eta $, so does the diagram
  
  We then have
  \begin{align*} \eta _{GF(B)}\circ GF(h) & = GFG(f)\circ \eta _{GF(A)}\\ & = \eta _{GF(B)}\circ G(f). \end{align*}

  Postcomposing with $\eta ^{-1}_{GF(B)}$ then gives $GF(h) = G(f)$. This shows $F$ to be full.

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  Essential surjectivity: Let $B\in \operatorname {\mathrm{Obj}}(\mathcal{D})$. We need to find some $A\in \operatorname {\mathrm{Obj}}(\mathcal{C})$ with $F(A)\cong B$. Choose $A\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}G(B)$. Then $\epsilon _{B}\colon FG(B)\to B$ is an isomorphism by assumption. Hence $F$ is essentially surjective.

Since $F$ is fully faithful, we can extend the assignment $B\mapsto j_{B}$ to a unique functor $j\colon \mathcal{D}\to \mathcal{C}$ such that the isomorphisms $i_{B}\colon B\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }F(j_{B})$ assemble into a natural isomorphism $\eta \colon \operatorname {\mathrm{id}}_{\mathcal{D}}\mathbin {\overset {\mathord {\sim }}{\Longrightarrow }}F\circ j$, with a similar natural isomorphism $\epsilon \colon \operatorname {\mathrm{id}}_{\mathcal{C}}\mathbin {\overset {\mathord {\sim }}{\Longrightarrow }}j\circ F$. Hence $F$ is an equivalence.

are natural isomorphisms iff $\epsilon $ and $\eta $ are.

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  When $F$ and $G$ Are Equivalences: By Item 5, if $F$ and $G$ are equivalences, then so is $G\circ F$.

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  When $G$ and $G\circ F$ Are Equivalences: If $G\circ F$ and $G$ are equivalences, note that by Item 5 and Item 3, so are $G^{-1}$ and $G^{-1}\circ G\circ F\cong F$. Therefore $F$ will be an equivalence by Item 2.

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  When $F$ and $G\circ F$ Are Equivalences: If $G\circ F$ and $F$ are equivalences, so are $F^{-1}$ and $G\circ F\circ F^{-1}\cong G$ by Item 5 and Item 3. Therefore $G$ will be an equivalence by Item 2.

This finishes the proof.

as the composition

as the composition

so that the components of $\eta ^{\circ }$ and $\epsilon ^{\circ }$ are of the form

Since $\eta ^{\circ }$ and $\epsilon ^{\circ }$ are compositions of whiskerings of natural isomorphisms, they are natural transformations by Definition 11.9.5.1.2 and natural isomorphisms by Item 5b and Item 5c of Item 5 of Proposition 11.9.5.1.3.

Thus, $(F'\circ F,G\circ G',\eta ^{\circ },\epsilon ^{\circ })$ is an equivalence.