Equivalence of categories is an equivalence relation Suppose that $B \xrightarrow{F}C$ and $C \xrightarrow{G}D$ are equivalences of categories. I want to show that $G \circ F$ is an equivalence.
(This becomes easy if I use that a functor is an equivalence of categories iff it is fully faithful and essentially surjective; however, I'm in the process of trying to prove this!)
Suppose that $F' \circ F \overset{\alpha}{\simeq} \text{Id}_B, \, F \circ F' \overset{\beta}{\simeq} \text{Id}_C,\, G' \circ G \overset{\gamma}{\simeq} \text{Id}_C,\, G \circ G' \overset{\delta}{\simeq} \text{Id}_D$. Now I want to show that $F' \circ G' \circ G \circ F \simeq \text{Id}_B$. What natural isomorphism should I use? 
I'm probably just overcomplicating this and missing something simple. Thanks!
 A: Cat is a 2-category. So you can compose natural transformations as 2-cells. The counit of the composite adjoint equivalence will be
$$F'G'GF \xrightarrow{1\gamma1} F'F \xrightarrow{\alpha} Id_B$$
and the unit will be defined in a similar way. This will work for adjoint functors as well. So in fact you get a category whose objects are categories and whose morphisms are adjoint pairs of functors.
A: I had issues solving this question algebraically so I am posting an algebraic solution for anyone else who might want to use it. I am showing only one way, but the other way around is the same:
Write
$$
\begin{array}{c}
\alpha:1_{\mathcal{B}}\widetilde{\Rightarrow}F^{\prime}F\\
\\
\gamma:1_{\mathcal{C}}\widetilde{\Rightarrow}G^{\prime}G
\end{array}
$$
For the Natural isomorphisms given. And define $$\zeta:1_{\mathcal{B}}\widetilde{\Rightarrow}F^{\prime}G^{\prime}GF$$
by $\zeta_{A}=F^{\prime}(\gamma_{F(A)})\circ\alpha_{A}$ for $A\in Ob(\mathcal {B})$. Finally
$$
\begin{array}{c}
F^{\prime}G^{\prime}GF(f)\circ\zeta_{A}=F^{\prime}\left(G^{\prime}GF(f)\right)\circ\zeta_{A}=\\
\\
F^{\prime}\left(\gamma_{F(B)}\circ F(f)\circ\gamma_{F(A)}^{-1}\right)\circ\zeta_{A}=\\
\\
F^{\prime}\left(\gamma_{F(B)}\right)\circ F^{\prime}F(f)\circ F^{\prime}\left(\gamma_{F(A)}^{-1}\right)\circ\zeta_{A}=\\
\\
F^{\prime}\left(\gamma_{F(B)}\right)\circ F^{\prime}F(f)\circ\underset{=Id}{\underbrace{F^{\prime}\left(\gamma_{F(A)}\right)^{-1}\circ F^{\prime}(\gamma_{F(A)})}}\circ\alpha_{A}=\\
\\
F^{\prime}\left(\gamma_{F(B)}\right)\circ F^{\prime}F(f)\circ\alpha_{A}=F^{\prime}\left(\gamma_{F(B)}\right)\circ\alpha_{B}\circ f=\zeta_{B}\circ f
\end{array}
$$
which shows $\zeta$ is indeed a Natural transformation. It is a Natural isomorphism since the composition of isomorphisms is an isomorphism, which finalize the question.
