Uniqueness of the Comparison Functor Suppose $F:C\rightarrow D$ and that $F\dashv U$ is an adjunction and $C^{T}$ the Eilenberg–Moore category for the monad $T=U◦F$, with the corresponding functors  $F^{T}:C\rightarrow C^{T}$ and $U^{T}:C^{T}\rightarrow C$. 
I have been able to prove that there is a comparison functor $Φ : D →C^{T}$ which satisfies 
(1) $U^{T}◦Φ= U$ 
and 
(2) $Φ◦F = F^{T}$ 
I am having trouble with uniqueness. 
Here is what I have so far: Suppose $Φ'$ satisfies (1) and (2).
Let $U\in D$. Then using (1) with $Φ'(D)=(C',\alpha )$, it follows that  $U◦Φ'(D)=U(C',\alpha )=C'$ whereas $Φ(D)=(UD,U\varepsilon_{D})$ and so $U◦Φ(D)=U(UD,\varepsilon_{D})=UD$ which says $C'=UD$
Now I need to show that $\alpha=U\varepsilon_{D}$. This is where I'm stuck.
edit: Using the hint below, the fact that the adjunctions have the same unit imply, after using (1) and (2) that
\begin{matrix}
\operatorname{Hom}(FC, D) & \xrightarrow{{\phi}} & \operatorname{Hom}(C, UD) \\
\left\downarrow\vphantom{\int}\right. & & \left\downarrow\vphantom{\int}\right.\\
\operatorname{Hom}(F^{T}C, Φ'(D))&  \xrightarrow{\phi^{T}} & \operatorname{Hom}(C, U^{T}Φ'(D))
\end{matrix}
commutes. 
($\phi$ and $\phi^{T}$ are the isomorphisms giving the adjunctions; the left downward arrow is the map $f\longmapstoΦ'(f)$ and the right downward arrow is the identity on the Hom$(C, UD)$.)
Then, setting $C=UD$ and following $id_{UD}$, you get that $Φ'\epsilon=\epsilon^T Φ'$, which is the hint. The rest follows easily. 
 A: Just a hint....
First you should realize that 
$$Φd=(Ud, h)$$
that is a T-algebra with underlying $Ud$, for any d in $D$. in order to obtain the structure h, observe that the 2 adjunctions have the same unit $\eta$, and deduce that $Φ\epsilon=\epsilon^T Φ$.
Then deduce that $\epsilon^TΦd=\epsilon^T(Ud, h)=h$ and so $Φ\epsilon=\epsilon^T Φ$ implies $U\epsilon_d=h$ so h is determined and $Φ$ is unique.
A: Here's a version of the proof that bypasses the $\Phi\epsilon=\epsilon^T\Phi$ lemma and proves uniqueness directly.


*

*$U^T\Phi=U$ tells us that $\Phi d$ is a $T$-algebra with structure map $\gamma d: TUd\to Ud$

*$\Phi$ sends $D$-arrows to $T$-homomorphisms, so $\gamma$ is a natural transformation $TU\to U$

*$\Phi F=F^T$ tells us that $\gamma F=\mu =U\epsilon F$


Since $\gamma$ is natural we have $U\epsilon\circ\mu U=U\epsilon\circ\gamma FU=\gamma\circ TU\epsilon$
Precomposing with $T\eta U$ gives us $U\epsilon\circ\mu U\circ T\eta U=\gamma\circ TU\epsilon\circ T\eta U$
This rearranges as $U\epsilon\circ(\mu\circ T\eta)U=\gamma\circ T(U\epsilon\circ\eta U)$, which simplifies to $U\epsilon=\gamma$ 
