Theorem: Equivalence of categories is an equivalence relation

Proof: Let $\mathbf{A}, \mathbf{B}$, and $\mathbf{C}$ be categories.

Clearly, the identity functor $\mathbf{A} \rightarrow \mathbf{A}$ is an equivalence of categories, so equivalence is reflexive.

Suppose $F : \mathbf{A} \rightarrow \mathbf{B}$ is an equivalence. Then we have $G : \mathbf{B} \rightarrow \mathbf{A}$ and natural isomorphisms $\eta : 1_{\mathbf{A}} \rightarrow G \circ F$ and $\varepsilon : F \circ G \rightarrow 1_{\mathbf{B}}$.

$$G$$ along with the natural isomorphisms $$\varepsilon^{-1} : 1_{\mathbf{B}} \rightarrow F \circ G$$ and $$\eta^{-1} : G \circ F \rightarrow 1_{\mathbf{A}}$$ is an equivalence from $$\mathbf{B}$$ to $$\mathbf{A}$$, so equivalence is symmetric. Finally, suppose $$F : \mathbf{A} \rightarrow \mathbf{B}$$ and $$X : \mathbf{B} \rightarrow \mathbf{C}$$ are equivalences. We have $$G : \mathbf{B} \rightarrow \mathbf{A}$$ and $$Y : \mathbf{C} \rightarrow \mathbf{B}$$ along with all of the following natural isomorphisms:$$

\eta^1 : 1{\mathbf{A}} \rightarrow GF \qquad \varepsilon^1 : FG \rightarrow 1{\mathbf{B}} \qquad \eta^2 : 1{\mathbf{B}} \rightarrow YX \qquad \varepsilon^2 : XY \rightarrow 1{\mathbf{C}}

$$

Consider $X \circ F : \mathbf{A} \rightarrow \mathbf{C}$, and $G \circ Y : \mathbf{C} \rightarrow \mathbf{A}$. We need to show that $GYXF : \mathbf{A} \rightarrow \mathbf{A}$ is naturally isomorphic to $1_{\mathbf{A}}$. We need to define $\eta^3 : 1_{\mathbf{A}} \rightarrow GYXF$. In fact, we will write it as the vertical composition of $\eta^1 : 1_{\mathbf{A}} \rightarrow GF$ and a new natural transformation $\gamma : GF \rightarrow GYXF$. To find an expression for $\gamma$, consider the following diagram:

We can then horizontally compose the natural transformations here (the identity natural transformation on the left and right), to yield a natural transformation $\gamma = G \eta^2 F : GF \rightarrow GYXF$ -- abusing notation a bit. To be clear, we define $\gamma_A = G(\eta^2_{F(A)})$.

Then, $\eta^3 = \gamma \circ \eta^1$ is a natural isomorphism from $1_{\mathbf{A}}$ to $GYXF$. We proceed similarly to find $\varepsilon^3 : XFGY \rightarrow 1_{\mathbf{C}}$. Consider:

so define the natural transformation $\varphi = X \varepsilon^1 Y : XFGY \rightarrow XY$. Then, let $\varepsilon^3 = \varepsilon^2 \circ \varphi$.

Finally, then, we have a pair of functors $\mathbf{A} \rightarrow \mathbf{C}$ and $\mathbf{C} \rightarrow \mathbf{A}$ along with $\eta^3$ and $\varepsilon^3$ that make the functors into an equivalence of categories, and such an equivalence is transitive.