Reflective and Coreflective subcategories give rise to idempotent functors. For a pair  $(\mathcal{X},\mathcal{Y})$ of full subcategories of  an abelian category $\mathcal{C}$ suppose the inclusion functor $i:\mathcal{X} \to \mathcal{C}$ has a right adjoint functor $R: \mathcal{C} \to \mathcal{X}$, here we are saying that $\mathcal{X}$ is a coreflective subcategory of $\mathcal{C}$ then the functor $iR:\mathcal{C} \to \mathcal{C}$ is idempotent.
What I need to prove is that for every object $C \in \mathcal{C}$ and every morphism $f \in \mathcal{C}$ we have that $(iR)^{2}=(iR)(iR)(C)=(iR)C$ and $(iR)^{2}=(iR)(iR)(f)=(iR)f$ but I'm pretty stuck proving this straightforward.
By duality we can also formulate that for the inclusion $j:\mathcal{Y} \to \mathcal{C}$ we have a left adjoint functor $L:\mathcal{Y} \to \mathcal{C}$ then $jL:\mathcal{C} \to \mathcal{C}$ is an idempotent functor also.
 A: You won't be able to prove this, the best you can get is $(iR)^2\cong iR$.
In fact, the point is that the counit $\epsilon : iR\to id$ becomes an isomorphism if you compose it with $iR$: $iR\epsilon : (iR)^2\to iR$ is an isomorphism.
This is not hard to prove. I'll do it for the dual case though, as I'm more used to it.
In fact, the better statement is that the counit $\epsilon: Lj\to id$ is an isomorphism.
The proof is easy : you have a map $LjY\to Y$ for any $Y\in \mathcal Y$, so by the Yoneda lemma, to prove that it's an isomorphism you only need to prove that for any $Y'\in\mathcal Y$, $\hom(Y,Y')\to\hom(LjY,Y')$ is an isomorphism.
But the latter is $\cong \hom(jY,jY')$ by adjunction, and you can easily check that the composite map $\hom(Y,Y')\to \hom(LjY,Y')\to \hom(jY,jY')$ is just application of the functor $j$.
But $j$ is fully faithful, so this composite map is an isomorphism: we are done.
(another way to phrase this is to use the fully-faithfulness to prove that $jY\to jY$ has the universal property of $jLjY\to jY$)
Since $\epsilon : Lj\to id$ is an isomorphism, you can use the triangle identity to prove that $L\eta$ and hence $jL\eta : jL\to (jL)^2$ is an isomorphism, which is the dual statement (in fact, $\eta jL$ too, will be an isomorphism)
A: Since $i:\mathcal X\hookrightarrow\mathcal C$ is a fully faithful functor, the unit of the adjunction $i\dashv R$ gives a natural isomorphism $\operatorname{Id}_{\mathcal X}\cong R\circ i$.
Consequently, $i\circ R\circ i\circ R\cong i\circ\operatorname{Id}_{\mathcal X}\circ R=i\circ R$.
Note that this holds of arbitrary categories.
