An ambient object lies in the essential image of a reflective inclusion exactly when its reflection unit is invertible

If ${\eta }_C$ is invertible, it itself displays $C$ as isomorphic to the included object $I(R(C))$, so $C$ lies in the essential image.

Conversely, let $u:C\to I(A)$ be an isomorphism. Naturality of $\eta$ gives $IR(u)\circ {\eta }_C={\eta }_{I(A)}\circ u$. The map $IR(u)$ is invertible because a functor sends the inverse of $u$ to its inverse. Applying $R⊣I$ in [L3] at $d=A$ gives $I({\epsilon }_A)\circ {\eta }_{I(A)}=1_{I(A)}$, and $I({\epsilon }_A)$ is invertible because ${\epsilon }_A$ is by [L1] and functors preserve inverses; composing that identity with $I({\epsilon }_A{)}^{-1}$ on the left gives ${\eta }_{I(A)}=I({\epsilon }_A{)}^{-1}$, which is therefore an isomorphism. Hence ${\eta }_C=IR(u{)}^{-1}\circ {\eta }_{I(A)}\circ u$. A composite $g\circ f$ of isomorphisms is an isomorphism, since $f^{-1}\circ g^{-1}$ is a two-sided inverse for it by associativity and the identity laws; applying this twice and using [L2] makes ${\eta }_C$ invertible.