I started the definition of adjoint functor using universal morphisms. A functor $F:\mathcal C\to \mathcal D$ is a left adjoint functor if given any object $Y\in D$, there is an object $G(Y)$ in $\mathcal D$ with a morphism $F(G(Y))\to Y$ such that for any $F(X)\to Y$, there is a unique morphism $X\to G(Y)$ such that the composition $F(X)\to F(G(Y))\to Y$ is the same as $F(X)\to Y$.
A functor $G:\mathcal D\to\mathcal C$ is a right adjoint functor if given any object $X\in C$, there is an object $F(X)$ in $\mathcal D$ and a morphism $X\to G(F(X))$ such that for any $X\to G(Y)$, there is a unique morphism $F(X)\to Y$ such that the composition $X\to G(F(X))\to G(Y)$ is the same as $X\to G(Y)$.
I am trying to verify only one direction, for example, if $F$ is a left adjoint functor with the $G$ described above, it's routine to verify $G$ is a functor, but how is $G$ is the right adjoint functor of $F$ in the sense of my second definition?
Pick any $X\in\mathcal C$, I am really looking for a universal morphism $X\to G(F(X))$ in the sense of my second definition. Should I use $Y=F(X)$ and apply the the first definition? Then I have a morphism $F(G(F(X)))\to F(X)$ such that for the identity $F(X)\to F(X)$, there is a unique morphism $X\to G(F(X))$ such that under $F$ factors through $F(G(F(X)))\to F(X)$. I was guessing then $X\to G(F(X))$ is the universal morphism I am looking for, but I wasn't able to prove it is universal, and the way I found it seems very unatural to me. What is a better, more clear way I can see this?
This is not exactly a short proof. I'm pretty sure that for instance Borceux's handbook has a detailed proof. Let me prove still the part of the proof I think is "harder" and give some guidelines.
It might be easier to go through the intermediate step of proving the triangular identities. i.e: $F$ is left adjoint to $G$ iff the triangular identities hold.
It is the direct proof "$F$ is left adjoint to $G$ implies the triangular identites" that I'm going to prove.
The converse is just a bunch of diagram-pastings.
Once you have this equivalence and you notice that the definition of triangular identities is autodual and that in the opposite category $\eta$ and $\varepsilon$ change their roles, you get the desired result by starting from $F$ being left adjoint to $G$, then you can apply to it the direct proof of the equivalence to get that $F$ and $G$ satisfy the triangular identities. Then you apply the converse of the equivalence to the opposite category (which satisfies the triangular identities by the observation) which gives you straight that $G$ is right adjoint to $F$.
For the proof I'm going to write $L : B → A$ instead of $F : C → D$ and $R : A → B$ instead of $G : D → C$.
Let $a,a':A$ and $f : a → a'$. We want to prove that the square at the bottom commutes, meaning that $\varepsilon$ is natural. (I completed the diagram above with the definition of adjointness for clarity.)
For this let's just use the definition of $L$ being left adjoint to $R$ in the following way:
This helps us in that by uniqueness, we know that it is enough to prove $R\ \varepsilon_{a'} ◦ (R\circ L◦R)\ f ◦ \eta_{R\ \! a} = R\ f \circ R\ \varepsilon_a ◦ \eta_{R\ \! a}$ to know that $\varepsilon_{a'} ◦ (L◦R)\ f = f \circ \varepsilon_a$, which is what we wanted.
To prove this identity, observe the following diagram:
Then the following series of transformations of diagram chasing proves the naturality of $\varepsilon$. (Remember that we already know that $\eta$ is natural by definition of $L$ being left adjoint to $R$.)
Now we want to prove the triangular identities themselves. I'm going to prove $L\ \varepsilon_{L\ \! b} \circ L\ \eta_b = id$ and the other one is similar.
Again we use the same trick and it's enough to prove that $R\ \varepsilon_{L\ \! b} ◦ (R◦L)\ \eta_b ◦ \eta_b = id ◦ \eta_b$:
To prove this we just have to paste the diagrams of the naturality of $\eta$ with the one of the reflection for $b:=(R◦L)\ b$ and $a:=L\ b$:
$\blacksquare$