TITLE: Reflective subcategory of a complete category
QUESTION [5 upvotes]: I am trying to understand why the reflective subcategory of a complete category is complete. Let $\mathcal{C}\subset \mathcal{D}$ with $\mathcal{D}$ complete and let $r:\mathcal{D}\to \mathcal{C}$ be the reflection. Consider the cone $(\lim X_k \to (X_k)_k)$ where all $X_k\in \mathcal{C}$. Each canonical map $\lim X_k \to X_k$ for each $k$ gives rise to the factorization $\lim X_k \to r(\lim X_k) \to X_k$ for any $k$, and since the factorization is unique, we get a new cone $(r(\lim X_k) \to (X_k)_k)$. Then each cone $(Y \to (X_k)_k)$ gives rise to the factorization $(Y \to \lim X_k \to r(\lim X_k) \to (X_k)_k)$. Why is the map $Y\to r(\lim X_k)$ unique ? I must use somewhere that the right adjoint of $r$ (the inclusion $\mathcal{C}\subset \mathcal{D}$) is full and faithful of course. Any help ? I cannot find the proof in the books I have with me. Thanks.

REPLY [3 votes]: You aren't just proving that $\mathcal C$ is complete, you know that it's complete in a specific way, ie. that limits are calculated the same way both in $\mathcal C$ and in $\mathcal D$, or in other words, that the canonical map $η : \lim X → r \lim X$ must be an isomorphism. (This is because the inclusion, being by assumption right adjoint, must preserve limits.)
If you prove that, the cone $(ρ_k : r(\lim X) → X_k)_k$ you already constructed will be isomorphic to the limiting cone $(π_k : \lim X → X_k)_k$, and so limiting itself.
Getting the inverse of $η$ is easy, because $ρ$ must factor through $π$ via a unique $\mathcal D$-morphism $ϑ$. From $π_kϑ = ρ_k$ and $ρ_kη = π_k$ one gets $π_kϑη = π_k$ for all $k$, and so $ϑη = \mathrm{id}$ by the universality of $π$. Since $\mathcal C$ is full, $ηϑ$ is a $C$-morphism, and by $ηϑη = η$ and the universality of $η$, it's the identity morphism too.
As a sidenote, a short non-elementary way to prove this is to note that the inclusions of full subcategories are monadic, and therefore create all limits.