Theorem: Let $F : \mathbf{A} \leftrightharpoons \mathbf{B} : G$ be functors and $\eta : 1_{\mathbf{A}} \rightarrow GF$, $\varepsilon : FG \rightarrow 1_{\mathbf{B}}$ natural isomorphisms such that we have an equivalence of categories between $\mathbf{A}$ and $\mathbf{B}$. Then $F$ is left adjoint to $G$.

Proof:

$$\eta$$ and $$\varepsilon$$ are certainly natural transformations, but they may not satisfy the triangle identities. In particular, we may have that $$\varepsilon F \circ F \eta \neq \id_F$$. Define a new natural transformation $$\varepsilon' : FG \rightarrow 1_{\mathbf{B}}$$. We will let $$\varepsilon' = \varepsilon \circ F \eta^{-1} G \circ FG \varepsilon^{-1}$$. Note that since $$\eta$$ and $$\epsilon$$ are natural isomorphisms, so is each member of the composition and thus $$\varepsilon'$$ as well. We claim that $$\varepsilon' F \circ F\eta = \id_F$$. Or, expanding out, that $$\varepsilon F \circ F \eta^{-1} G F \circ FG \varepsilon^{-1} F \circ F \eta = \id_F$$. Let $$A$$ be arbitrary and examine the component of the left-hand side at $$A$$:$$

FA \xrightarrow{F(\etaA)} FGFA \xrightarrow{FG(\varepsilon^{-1}{FA})} FGFGFA \xrightarrow{F(\eta^{-1}{GFA})} FGFA \xrightarrow{\varepsilon{FA}} FA

$$we want this to be equal to $$1_{FA}$$. First, we write down some commutative diagrams that give some useful identities about the components of $$\eta$$ and $$\varepsilon$$ that depend on the fact that they are isomorphisms. First, note that this square commutes: <div class="math-figure"><img src="/assets/math_solutions/leinster/e2-3-10_1.svg" width="300px"/></div> since is is the statement of the naturality requirement for the natural isomorphism $$\eta : 1 \rightarrow GF$$ with respect to the arrow $$\eta_A : A \rightarrow GFA$$. This diagram states that $$\eta_{GFA} \circ \eta_A = GF(\eta_A) \circ \eta_A$$. Since $$\eta_A$$ is an isomorphism, we can compose on the right by $$\eta_A^{-1}$$ to get$$

\eta_{GFA} = GF(\eta_A)

$$Pull the dual trick with $$\varepsilon$$. That is, since this diagram commutes: <div class="math-figure"><img src="/assets/math_solutions/leinster/e2-3-10_2.svg" width="300px"/></div> we can say that $$\varepsilon_B \circ FG(\varepsilon_B) = \varepsilon_B \circ \varepsilon_{FGB}$$, and left-cancel to get:$$

FG(\varepsilonB) = \varepsilon{FGB}

$$

Now, we write down another commutative diagram:

This one commutes because it is precisely the statement of the naturality for the natural isomorphism $\varepsilon^{-1} : 1 \rightarrow FG$ with respect to the arrow $F(\eta_A) : FA \rightarrow FGFA$. Applying the identities we just derived (keeping in mind they also work for the inverses of $\eta$ and $\varepsilon$), we can rewrite the right and bottom arrows.

Finally, note that the bottom arrow is an isomorphism and has an inverse $F(\eta_{GFA})^{-1} = F(\eta^{-1}_{GFA})$. With this, simply attach an $\varepsilon_{FA} : FGFA \rightarrow FA$ to the bottom of the diagram to get the final commuting diagram:

Following the diagram right, down, left, and down gives exactly the composite in question from earlier. But, following the left side straight down gives $\varepsilon_{FA} \circ \varepsilon^{-1}_{FA} = 1_{FA}$. Since the diagram commutes, we have the desired equality.

So, $\eta$ and $\varepsilon'$ are natural isomorphisms satisfying the triangle inequalities. So $F \dashv G$ with unit $\eta$ and counit $\varepsilon'$.