If $\mathcal{D}$ is reflective and $\mathcal{C}$ is complete, then $\mathcal{D}$ is complete.

Question

Let $\mathcal{D}$ be a sub-category of $\mathcal{C}$. We say it is reflective if:

1. $\mathcal{D}$ is full

2. the inclusion $i: \mathcal{D} \to \mathcal{C}$ admits a left adjoint $F$.

3. if $x \in \mathcal{C}$ is isomorphic to $y \in \mathcal{D}$ then $x \in \mathcal{D}$ too.

I have shown (without using condition 3) that if $\mathcal{D}$ is reflective and $\mathcal{C}$ is co-complete, then $\mathcal{D}$ is co-complete as well.

I'm asked as a follow up to prove:

If $\mathcal{D}$ is reflective and $\mathcal{C}$ is complete, then $\mathcal{D}$ is complete.

I know this can probably be proved directly in a similar way for the co-complete assertion, however I want to prove this using what I already have. I was thinking that if $\mathcal{C}$ is complete, than $\mathcal{C}^{op}$ is co-complete. If I can prove that $\mathcal{D}^{op}$ is reflective I'd be done. But this doesn't seem to work because I can't guarantee the inclusion has a left adjoint..

Questions:

1. How can I prove this assertion using the dual assertion on co-complete categories.

2. I'm also wondering if I've done something wrong in proving the co-complete assertion; is condition 3 required for this at all?

Answer 1

Here are a couple of thoughts :

First of all, you're right that 3. is useless to prove these things, it's probably just a convenience for the authors or your professor. I will however use it to simplify how to write things : it is indeed convenient.

Secondly, as I mentioned in the comments, a subcategory can be reflective without being co-reflective (i.e. its opposite is not necessarily reflective) therefore you cannot apply the previous result that easily (if you can apply it at all)

Lastly, here's an idea of how a proof might go :

Let $I$ be a small category and $X: I\to \mathcal{D}$ a functor. Compose with $i$ and let $(L, (\pi_j)_{j\in I})$ be the limit in $\mathcal{C}$.

Now recall the following fact : if $i:\mathcal{D\to C}$ is a reflective subcategory; then for $x\in \mathcal{C}$, $x\in \mathcal{D}$ (that's where 3. comes into play : if you don't have 3. you have to say "$x$ is isomorphic to some object of $\mathcal{D}$", but fundamentally nothing changes as limits are invariant under isomorphism) if and only if the unit of the adjunction $\eta_x : x\to iF (x)$ is an isomorphism. If you don't know that fact, prove it !

Therefore if we can prove that $\eta_L : L\to iF(L)$ is an isomorphism, then $L\in \mathcal{D}$ and since $\mathcal{D}$ is full, $(L,(\pi_j)_{j\in I})$ will also be a limit in $\mathcal{D}$.

Now note the following : $iF(L)\to^{iF(\pi_j)} iF(X_j) \to X_j$ is also a cone, where the second morphism is the inverse of $\eta_{X_j}$ (which exists, by the fact I mentioned earlier), hence we have a map of cones $f:iF(L)\to L$ through which it factors; and the map $L\to^{\eta_L} iF(L)$ which you can easily see is also a map of cones ($\eta$ is natural).

Therefore when you compose the two you get a map of cones $L\to L$, which must therefore be the identity. The only thing left to prove is that the composite in the other direction is also the identity : $\eta_L \circ f : iF(L)\to iF(L)$

Now there's another general fact that says we're actually done, which I will prove : if $i:\mathcal{D\to C}$ is a reflective subcategory, and $x\in\mathcal{C}$, then for $x$ to be in $\mathcal{D}$, it suffices that there is a retraction $iF(x)\to x$.

Proof : let $f$ denote the retraction (to be in keeping with the previous notations), and consider the naturality diagram for the adjunction for the map $(\eta_L^{op}, id)$, that is

$\require{AMScd} \begin{CD} \hom(iF(x),iF(x)) @>>> \hom(FiF(x), F(x))\\ @V{-\circ \eta_x}VV @VV{-\circ F(\eta_x)}V\\ \hom(x,iF(x)) @>>> \hom(F(x),F(x)) \end{CD}$

Start from $\eta_x\circ f$ in the top left hand corner, then going down yields $\eta_x\circ f\circ \eta_x= \eta_x$ (because $f$ is a retraction), which is then sent to $id_{F(x)}$ by definition of $\eta_x$. Now the point is that the diagonal is actually the "identity".

Indeed, consider the two maps $F(x)\to^{F(\eta_x)} FiF(x), FiF(x)\to^{\epsilon_{F(x)}} F(x)$. The second one is known to be an isomorpism because $\mathcal{D}$ is $F\dashv i$ is specifically a reflection-adjunction so $\epsilon$ is always an isomorphism, and the triangle identities tell you that when you compose them you get the identity $F(x)\to F(x)$. It follows that $F(\eta_x)$ is precisely the inverse of $\epsilon_{F(x)}$ so if you start at the bottom right hand corner with $g$, you get up to $g\circ \epsilon_{F(x)}$, and then you go left to $i(g\circ \epsilon_{F(x)}) \circ \eta_{iF(x)}$ and then again, the triangle identities tell you that this is nothing but $i(g)$, i.e. "$g$". Therefore the diagonal map is $i(g)\to g$ from which it follows, since $\eta_x\circ f \mapsto id_{F(x)}$, that $\eta_x\circ f = id_{iF(x)}$ and so the retraction was actally an isomorphism.

It's very likely that there's a simpler proof, I just didn't have any paper at hand to write it down so it was more complicated to see what's going on.

If you think about it in terms of preorders it's pretty clear what's going on : a retraction tells us that $iF(x) \leq x$ but we already know $x\leq iF(x)$