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Let $\mathcal{A}\subset \mathcal{B}$ be two categories with $\mathcal{A}$ full and reflective in $\mathcal{B}$. Let $R:\mathcal{B}\to\mathcal{A}$ be the reflection. That $R$ is the left adjoint to the inclusion $\mathcal{A}\subset \mathcal{B}$ is equivalent to the fact that every object of $\mathcal{A}$ is orthogonal to all units $X\to R(X)$ for $X$ running over the class of objects of $\mathcal{B}$. Conversely, let $Z$ be an object of $\mathcal{B}$ which is orthogonal to all maps $X\to R(X)$ for $X$ running over the class of objects of $\mathcal{B}$.

If $\mathcal{A}$ and $\mathcal{B}$ are locally presentable, then the inclusion $\mathcal{A}\subset \mathcal{B}$ is $\lambda$-accessible for some regular cardinal $\lambda$. Choose $\lambda$ large enough so that $\mathcal{A}$ and $\mathcal{B}$ are locally $\lambda$-presentable. Then $\mathcal{A}$ is a $\lambda$-orthogonality class of $\mathcal{B}$. Which implies that there exists a set of $\lambda$-presentable objects of $\mathcal{B}$ such that the objects of $\mathcal{A}$ are the objects of $\mathcal{B}$ which are orthogonal to the maps $Y\to R(Y)$ for $Y$ running over this set. In particular, this implies that $Z$ belongs to $\mathcal{A}$.

Can we conclude that $Z$ belongs to $\mathcal{A}$ in full generality, without the local presentability hypothesis ? (I guess not.) I ask the question in case I am missing something. In my case, I work with locally presentable categories.

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Since $Z$ is right orthogonal to every component of the adjunction unit, it is in particular right orthogonal to $\eta_Z : Z \to R Z$. Thus $\eta_Z : Z \to R Z$ admits a retraction, say $r : R Z \to Z$. Let $a = \eta_Z \circ r$. It is an idempotent endomorphism of $R Z$. But $\eta_Z \circ r = R r \circ \eta_{R Z}$, and both $\eta_{R Z} : R Z \to R R Z$ and $R r : R R Z \to R Z$ are isomorphisms, so $a$ is an idempotent automorphism of $R Z$, i.e. $a = \mathrm{id}_Z$. That implies $\eta_Z : Z \to R Z$ is an isomorphism.

We may therefore conclude that $Z$ is isomorphic to an object in the reflective subcategory. Moreover, if the reflective subcategory is replete, then $Z$ itself is in the reflective subcategory.

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  • $\begingroup$ I use isomorphism-closed instead of replete. The subcategory was supposed to be replete indeed. Thanks for the answer. $\endgroup$ Mar 26, 2014 at 9:57

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