$\newcommand{\lim}{\mathsf{lim}}$
# Weighted Limits in Reflective Subcategories

Reflective subcategories are closed under weighted limits.

(This is a result from [Kelly's Basic Concepts of Enriched Category Theory](http://www.tac.mta.ca/tac/reprints/articles/10/tr10.pdf), Chapter 3.5.)

In more detail: Assume that
1. $\mathcal V$ is a complete and cocomplete symmetric closed monoidal category,
2. $L\dashv I:\mathcal B\to \mathcal C$ is a reflective subcategory of $\mathcal V$-categories with unit $\eta:Id\to IL$.
3. $\phi:\mathcal J\to\mathcal V$ and $D:\mathcal J\to\mathcal B$ are $\mathcal V$-functors.

Then, if $\lim_\phi ID$ exists in $\mathcal C$, it is isomorphic to $\lim_\phi D$ in $\mathcal B$.

**Remark:** $I$ is faithful, because we say subcategory. Following current practice (see eg Borceux and the nLab), but not Mac Lane, we include in "reflective" also the condition that $I$ is full. A classical result, see Theorem 1 in Mac Lane, Chapter IV.3, says that for any adjunction the counit is iso iff the right adjoint is full and faithful. Therefore, item (2) implies that $LI\cong Id$. In other words, $IL$ is a so-called idempotent monad. These monads are similar to closure operators. In particular $ILIL\cong IL$. Finally, following again Borceux, we will assume that reflective subcategories are closed under isomorphisms.

**Notation:** From now on, we will drop "$\mathcal V$-" in statements like (2) and (3) above.

## Reflective subcategories are closed under retracts

The proof of the theorem is illuminated by a special case:

**Proposition:** (Kelly, Chapter 1.11) Reflective subcategories are closed under retracts.

*Proof:* Let $C\stackrel{s}{\longrightarrow} IB\stackrel{r}{\longrightarrow} C$ be a retract of $IB$, that is, $r\circ s=id$. By assumption, $\eta_{IB}$ is an iso. We have to show that $\eta_{C}$ is an iso. But a straightforward diagram chase reveals that $r\circ\eta_{IB}^{-1}\circ ILs$ is the inverse of $\eta_C$. QED

**Remark:** The notion of retract is one from ordinary category theory, so a more detailed statement of the proposition would read as follows. If $\mathcal B\to \mathcal C$ is a full reflective subcategory then $\mathcal B_0$ is closed under retracts in $\mathcal C_0$, where $(-)_0$ is the forgetful functor form $\mathcal V$-categories to oridinary categories.

## Reflective subcategories are closed under limits

Before we state the theorem recall [weighted limits](https://hackmd.io/@alexhkurz/BJwOsVWSL).

Despite left-adjoints not preserving limits in general, in the special case above we do have $L(\lim_\phi ID)\cong \lim_\phi LID$:

**Theorem:** If $\lim_\phi ID$ exists in $\mathcal C$ then $\lim_\phi D$ exists in $\mathcal B$ and $L(\lim_\phi ID)\cong \lim_\phi D$.

*Proof:* We have isomorphisms natural in $C$
\begin{align*}
\mathcal C(C,\lim_\phi ID)
&\cong
[\mathcal I,\mathcal V](\phi,\mathcal C(C,ID))
\\
& \cong [\mathcal I,\mathcal V](\phi,\mathcal C(ILC,ID))
\end{align*}

where the first one is the def of $\lim$ and the second one comes from
$$\mathcal C(\eta_C,1):\mathcal C(ILC,ID)\to \mathcal C(C,ID)$$

where $\eta_C:C\to ILC$ is the unit of the adjunction of Assumption (2). That $\mathcal C(\eta_C,1)$ is an iso follows from the adjunction $L\dashv I$ and $I$ being full and faithful.

The first iso, with $C=\lim_\phi ID$, yields a limiting cylinder
$$\mu:\phi\to\mathcal C(\lim_\phi ID,ID)$$

and composing with $\mathcal C(\eta_C,1)^{-1}$ gives us a cylinder
$$\alpha:\phi\to \mathcal C(IL\lim_\phi ID,ID)$$

which must be [^representable] of the form
$$\alpha = \mathcal C(f,ID)\circ \mu$$

for a unique $f:IL(\lim_\phi ID) \to\lim_\phi ID$. By uniqueness of representations, it now follows that $\lim_\phi ID$ is a retract of $IL(\lim_\phi ID)$, that is,
$$\lim_\phi ID \stackrel\eta\longrightarrow IL(\lim_\phi ID) \stackrel f \longrightarrow \lim_\phi ID \ = \ id $$

(An application of the proposition now shows that $\lim_\phi ID$ is in $\mathcal B$.) Finally, using that $I$ is full [^full], it follows that
$$IL\lim_\phi ID \stackrel f \longrightarrow \lim_\phi ID
\stackrel\eta\longrightarrow IL(\lim_\phi ID) \ = \ id $$

and, in particular $IL(\lim_\phi ID)\cong \lim_\phi ID$.

**Remark:** The proof above is an enriched version of the ordinary proof, see eg Borceux, Handbook of Categorical Algebra, Vol 1, Proposition 3.5.3: One first shows that $IL(\lim D)\to D$ is a cone over $D$ and that, therefore, there is $f:IL(\lim D)\to\lim D$. Then one shows that $\lim D$ is a retract of $IL\lim D$:
$$\lim D\stackrel \eta \longrightarrow IL\lim D \stackrel f\longrightarrow \lim D \ = \ id$$

It is instructive to compare Kelly's proof (the one we reproduced here) with Borceux's and to see how the former replaces the use of diagrams by arguments based on representability.

**Remark:** Another proof uses that $\mathcal B$ is the category of algebras for the monad $IL$ and that $I:\mathcal B\to\mathcal C$ is monadic and that monadic functors create limits.

## References

[Kelly: Basic Concepts of Enriched Category Theory](http://www.tac.mta.ca/tac/reprints/articles/10/tr10.pdf) (1982). Chapter 3.5

Borceux: Handbook of Categorical Algebra (1994). Chapter 3.5.

Mac Lane: Categories for the Working Mathematician (1972). Chapter IV.3.

nLab: [reflective subcategory](https://ncatlab.org/nlab/show/reflective+subcategory) and [weighted limit](https://ncatlab.org/nlab/show/weighted+limit).

[^representable]: The functor $[\mathcal I,\mathcal V](\phi,\mathcal C(-,ID))$ is represented by $\mathcal C(-,\lim_\phi ID)$. Hence, by a general fact about representable functors (see also the note on [limiting cylinders](https://hackmd.io/@alexhkurz/BJwOsVWSL#The-limiting-cylinder-of-a-weighted-limit)), we know that for all $\alpha:\phi\to \mathcal C(ILC,ID)$ there must be $f:ILC \to\lim_\phi ID$ such that $\alpha$ factors through the counit (Kelly, Chapter 1.10) $\mu$ of the representation, in symbols, $\alpha=\mathcal C(f,ID)\circ \mu$.

[^full]: This seems to be the only place where we use that $I$ is full. (Check again) That would mean that closure under limits does not depend on fullness of $I$. Can that be true? Should be known somewhere ...