Idempotent and adjoint Good day to everyone,
I have the next doubt about this problem: "Given the functors $G:A\rightarrow X$ and $K:X\rightarrow A$ and natural transformations $\alpha:KG\rightarrow I_{A}$, $\varrho:I_{X}\rightarrow GK$ such that $G\alpha\cdot\varrho G=1_{G}:G\rightarrow GKG\rightarrow G$.
Prove that $\alpha K\cdot K\varrho$ is and idempotent in $A^{X}$ (Category of functors of X to A) and $G$ has a left adjoint if and only if  $\alpha K\cdot K\varrho$ splits "
The idea for the first partis that I define:
$\varphi:A(Kx,a)\rightarrow X(x,Ga)$ by $\varphi(h)=G(h)\circ\varrho_{x}$
and
$\psi:X(x,Ga)\rightarrow A(Kx,a)$ by $\psi(f)=\alpha_{a}\circ K(f)$
I am going to have $\varphi\circ\psi=1$ with this i have that $\psi\circ\varphi$ is idempotent and it follows that $\alpha K\cdot K\varrho$ is an idempotent.
For the second part, if $G$ has a left adjoint then we have the next natural isomorphism
$\phi:A(Fx,a)\rightarrow X(x,Ga)$ and natural transformations $\epsilon:FG\rightarrow I_{A}$  and $\eta:I_{X}\rightarrow GF$, for $a=Kx$, I define $\theta_{x}=\phi^{-1}(\varrho_{x})=\epsilon_{Kx}\circ F\varrho_{x}$ for all $x\in Obj(X) $.
I mean $\theta=\epsilon_{K}\cdot F\varrho:F\rightarrow K$ is a natural transformation.
For $\psi:X(x,Ga)\rightarrow A(Kx,a)$, I set $a=Kx$, i define $\beta_{x}=\psi(\eta_{x})=\alpha_{Fx}\circ K\eta_{x}$ 
I mean $\beta=\alpha_{F}\cdot K\eta:K\rightarrow F$ is another natural transformation.
I am lost when i want to prove that $\alpha K\cdot K\varrho=\theta\cdot\beta$ and $\beta\cdot\theta=1$, I don't if that is correct, any suggestion will be well received and thank you for your time!!!

This is not a complete answer only how to prove one of the specific identities you asked about.
Since it is difficult to draw diagrams on here, I will write it as an equation:
$(\theta \cdot \beta  )_{x} = \epsilon_{K(x)} F(\rho_x) \alpha_{F(x)} K(\eta_x)\\
\phantom{(\theta \cdot \beta  )_{x}} = \alpha_{K(x)} KG(\epsilon_{K(x)}F(\rho_X))K(\eta_X)\qquad \text{by naturality of $\alpha$}\\
\phantom{(\theta \cdot \beta  )_{x}} = \alpha_{K(x)} KG(\epsilon_{K(x)})K(GF(\rho_X)\eta_X)\\
\phantom{(\theta \cdot \beta  )_{x}} = \alpha_{K(x)} KG(\epsilon_{K(x)})K( \eta_{GK(x)}\rho_X)\qquad \text{by naturality of $\eta$}\\
\phantom{(\theta \cdot \beta  )_{x}} = \alpha_{K(x)} K(G(\epsilon_{K(x)})\eta_{GK(x)})K(\rho_X )\\
\phantom{(\theta \cdot \beta  )_{x}} = \alpha_{K(x)} K(\rho_X )\qquad \qquad \qquad\qquad\text{by the triangular identity}\\
\phantom{(\theta \cdot \beta  )_{x}} = ((\alpha K)\cdot (K\rho))_x$
The other identity is obtained in a similar way by swapping $\eta$ and $\rho$, and $\epsilon$ and $\alpha$.
