# Counit component $\epsilon_d$ is an isomorphism only if $d$ lies in the essential image of the left adjoint

Let $$L\dashv R$$ be a pair of adjoint functors (say $$L\colon C\to D$$ and $$R\colon D\to C$$) with unit $$\eta\colon \mathrm{Id}\to R\circ L$$ and counit $$\epsilon\colon L\circ R\to \mathrm{Id}$$. Suppose that $$L$$ is fully faithful (hence that the unit is always an isomorphism).

Question: Does it follow that for each $$d\in D$$, the counit component $$\epsilon_d\colon L(R(d))\to \mathrm{id}$$ is an isomorphism if and only if $$d$$ lies in the essential image of $$L$$?

If $$\epsilon_d$$ is an isomorphism, then clearly $$d$$ lies in the essential image, but I'm not sure about the converse.

The motivation for the question is the idea that each adjunction restricts to an equivalence of subcategories (the subcategories given by the objects for which the unit or counit is an isomorphism). If the left adjoint happens to be fully faithful, it would be conceptually nice to say that the subcategory $$D$$ one gets in this way is precisely the essential image of $$L$$.

Example: Consider the adjunction between any category $$C$$ and the category of simplicial objects of $$C$$ given by the evaluation at $$[0]$$ and the "constant simplicial object" functor. The left adjoint ("constant simplicial object") happens to be fully faithful.

• Here's a proof of the dual. Dec 23, 2023 at 10:18

Yes, this is a simple consequence of the triangle identities: if $$d\simeq L(c)$$, then $$L(c)\to LRL(c)\to L(c)$$ is the identity, where the first map is $$L(\eta_c)$$ and the second map is $$\epsilon_{L(c)}$$. Since the first map is an isomorphism, it follows that the second one must be an isomorphism too.