Uniqueness of the Comparison Functor

Question

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.

Answer 1

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 $U\epsilon=\gamma$

Answer 2

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.