# Given an adjunction $\langle F,G,\eta, \epsilon\rangle$ with either $F$ or $G$ full, prove that $G\epsilon$ is invertible

I have this doubt with this problem:

"Given an adjunction $\langle F,G,\eta, \epsilon\rangle$ with either $F$ or $G$ full, prove that $G\epsilon:GFG\rightarrow G$ is invertible with inverse $\eta_{G}:G\rightarrow GFG$."

If $G$ is full the idea is $\eta_{Ga}=G(h)$ for some $h:a\rightarrow FGa$ because $G$ is full, then by naturality of $\eta$ we have:

$\begin{eqnarray} \eta_{Ga}\circ G\epsilon_{a}&=&G(h)\circ G\epsilon_{a}\\ &=&G(h\circ\epsilon_{a})\\ &=&G(\epsilon_{FGa}\circ FG(h))\\ &=&G(\epsilon_{FGA}\circ F\eta_{Ga})\\ &=&G(id)\\ &=&id \end{eqnarray}$

By the triangular identity we have $G\epsilon_{a}\circ \eta_{Ga}=id$, therefore $G\epsilon$ is invertible (if you find a mistake i will appreciate it).

My doubt is if $F$, I do not how to start from that. I think i do not see something in that. Any idea will be of alot of help.

Thank you!!!

A basic fact is that given an adjunction $\langle F,G,\eta,\epsilon\rangle:\mathcal{A}\to\mathcal{X}$. The functor $G$ is full if and only if each component of $\epsilon$ is a split monomorphism (see e.g. Mac Lane's Categories for the working mathematician). The dual of this is that $F$ is full if and only if each component of $\eta$ is a split epimorphism. Now since for each $a$ we always have $$G(\epsilon_a)\eta_{G(a)}=1_{G(a)}.$$ In the first case $G(\epsilon_a)$ is both a split epimorphism and a split monomorphism (since split monomorphisms are preserved by functors) and hence an isomorphism with inverse $\eta_{G(a)}$. While in the second case $\eta_{G(a)}$ is both a split epimorphism and a split monomorphism and hence an isomorphism with inverse $G(\epsilon_a)$.

Let us prove the basic fact above (our proof will follow the same lines as the one in Mac Lane's Categories for the working mathematician). To do so it is convenient to use the following lemma.

Lemma: For any morphism $p:E\to B$. Each component of the natural transformation $\hom(-,p):\hom(-,E)\to \hom(-,B)$ is an epimorphism if and only if $p:E\to B$ is a split epimorphism.

Proof: Trivially if $p$ is a split epimorphism then so will each component of $\hom(-,p)$. Conversely since $\hom(B,p):\hom(B,E)\to\hom(B,B)$ is an epimorphism, there exists $s:B\to E$ such that $\hom(B,p)(s)=1_B$ that is $ps=1_B$.

Now given an adjunction $\langle F,G,\eta,\epsilon:\mathcal X\to\mathcal A$ we see that for each $x$ and $y$ in $\mathcal X$ the composite of $F_{x,y}:\hom(x,y)\to\hom(F(x),F(y))$ and the isomorphism $\hom(F(x),F(y))\to\hom(x,GF(y))$ obtained from the adjunction is the map $\hom(x,\eta_y)$. This means $F$ is full (i.e. the maps $F_{x,y}$ are epimorphisms) if and only if the maps $\hom(x,\eta_y)$ are epimorphisms if and only if the morphisms $\eta_y$ are split epimorphisms (according to the previous Lemma).

Note 1 It seems worth mentioning that the above lemma can be phrased in terms of the Yoneda embedding $y:\mathcal C \to \mathbf{Set}^{C^{op}}$ and says for $p:E\to B$ in $\mathcal C$ the morphism $y(p)$ is an epimorphism if and only if $p$ is a split epimorphism.

• Thank you for the idea!!! That helped alot however I am gonna read that proposition because I want to understand the idea 😊 – Liddo Dec 1 '15 at 13:03
• I can put it here if you want. – Nex Dec 1 '15 at 15:02
• if you don't mind however it is not necessary, it is up to you!!! – Liddo Dec 1 '15 at 18:29