# The comparison functor for the adjoint $F\dashv G$ with $G$ fully faithful is an equivalence

Let $\mathcal{D} \underset{G}{\overset{F}{\leftrightarrows}}\mathcal{C}$ with $F\dashv G$ and $G$ fully faithful. Write $T=GF$. $\epsilon$ denotes the counit of the adjunction.
The comparison functor $K:\mathcal{D}\rightarrow TAlg$ for this adjunction is given by $K(D)=(GD,G\epsilon_D)$ and $K(f:A\rightarrow B) = G(f):G(A)\rightarrow G(B)$.
I want to prove that $K$ is an equivalence.

My attempt:
If I prove that $K$ is fully faithful and essentially surjective on objects, I'm done.
It was easy to see that $K$ inherits the fullness property of $G$. Similarly, using faithfulness of $G$ I was able to prove straight away that $K$ is faithful.
To prove that $K$ is essentially surjective: Let $(X,h)$ be a T-algebra. We want to show that there is $D\in\mathcal{D}$ such that $K(D)\overset{\phi}{\cong} (X,h)$.
Well, we can write $h:T(X)\rightarrow X$, so setting $D=T(X)$ and $\phi=h$ seems like a good bet. But I don't see why $h$ should be an isomorphism. I was thinking on trying to find $h^{-1}$ using the fullness of $G$, but failed.

Any hints?

• If $(X,h)$ is a $T$-algebra, then $h\eta_X=1$. So to prove that $h$ is an iso it is enough prove that $\eta_Xh=1$ as well. Apr 5, 2018 at 15:43
• @ArnaudD. Ok, great! Then I did the following: $\eta_X h = \mu_X T\eta_X=(..easy..)=id_{TX}$ where in the first equality I use that $\eta_X$ is a morphism of T-algebras. But I am not sure why this is true..
– Soap
Apr 5, 2018 at 17:39
• Well, in general it's not true... Use instead the fact that if $\mu$ is an isomorphism, then $T(\eta)=\eta_T$ to prove that $h$ must be an iso in $\mathcal{C}$. Apr 5, 2018 at 18:30
• @ArnaudD. Yes, I was able to show that $\mu$ is an isomorphism and that this implies $T\eta=\eta_T$. Now I've been trying to use that to show that $\eta_X h =1$ but I don't see how
– Soap
Apr 6, 2018 at 15:39
• Ok, new hint : use the naturality of $\eta$. Apr 6, 2018 at 16:04

If the right adjoint $G$ is fully faithful, then the counit $\epsilon$ is an isomorphism, and as a consequence, so is the multiplication $\mu=G(\epsilon_F)$, which means that the monad $T$ is idempotent. Since $\mu T(\eta)=1=\mu\eta_T$, we then have $T(\eta)=\eta_T=\mu^{-1}$.
Now an algebra for an idempotent monad is the same thing as an object $X$ such that $\eta_X$ is an isomorphism. Indeed, if $(X,h:TX\to X)$ is an algebra, then $h\eta_X=1$, and moreover $$\eta_Xh=T(h)\eta_{TX}=T(h)T(\eta_X)=T(h\eta_X)=1$$ since $\eta:Id\Rightarrow T$ is a natural transformation. Conversely, if $\eta_X$ is an isomorphism, then $h=\eta_X^{-1}:TX\to X$ makes $X$ an algebra, since $h\eta_X=1$ holds trivially and $$h T(h)=hT(\eta_X^{-1})=hT(\eta_X)^{-1}=h\mu.$$ Moreover, any map $f:X\to Y$ between two objects such that $\eta_X$ and $\eta_Y$ are isomorphisms is an algebra map, since we always have $\eta_Yf=T(f)\eta_X$ and thus also $f\eta_X^{-1}=\eta_Y^{-1}T(f)$.
So the category of $T$-algebra is equivalent (and even isomorphic) to the full subcategory $\mathcal{C}'$ of objects $X$ such that $\eta_X$ is an isomorphism. Since $\eta_{FY}=F(\epsilon_Y)^{-1}$ for all object $Y$ of $\mathcal{D}$ (by the triangle identity), the original adjunction restricts to an adjunction $\mathcal{D} \underset{G}{\overset{F}{\leftrightarrows}}\mathcal{C}'$, where now all components of the unit and counit are isomorphisms. Thus $G$ restricts to an equivalence between $\mathcal{D}$ and $\mathcal{C}'$, which is in fact the comparison functor $\mathcal{D}\to \mathcal{C}^T$.
There's an almost-one liner for that: the counit of $F\dashv G$ is an isomorphism, since $G$ is fully faithful, so $FG\cong 1$, and $K$ is such that $U^TK=G$, $KF=F^T$ if $F^T\dashv U^T$ denotes the free-forgetful adjunction generated by $T$.
If you put these things, together, it's pretty easy to show that $FU^T$ is an inverse for $K$:
1. $FU^TK = FG\cong 1$, because $G$ is fully faithful.
2. $KFU^T=F^T U^T \cong 1$, because the monad $T$ is idempotent ($T$ is idempotent iff $U^T$ is fully faithful, if and only if the counit $\epsilon^T : F^TU^T\Rightarrow 1$ is invertible, if and only if the multiplication of $T$ is an isomorphism).