Is the Yoneda completion of the rationals the extended real line? A thing that I have often heard is that, when viewed as an enriched category, the Yoneda embedding of the poset of rationals $\mathbb{Q}$ into its category of presheaves is just the dense embedding of the rationals into their completion, the extended real line. So the Dedekind completion of the rationals is a special case of the Yoneda embedding of a category into its free cocompletion. For example this fact is mentioned in a very old (early 2012) blog post by Qiaochu on the Yoneda lemma.
But the category of presheaves on $\mathbb{Q}$ is just the set of downward closed sets. And for every rational $r$ there are two downward sets: $(-\infty,r)=\{x|x < r\}$ and $(-\infty,r]=\{x|x \leq r\}.$ The latter presheaf is the representable presheaf, whereas the former is not representable, it's a new presheaf added by the Yoneda embedding.
So it appears that the category of presheaves actually contains two copies of every rational number, right next to each other one less than the other, along with only one of every irrational.
This does not appear to be order isomorphic to the extended reals. How do we make this work out? In a general category, for its free cocompletion via the Yoneda embedding, does the presheaf category contain extra copies of colimits that already existed in the starting category?
 A: You're right; this is an error in what you've heard. The embedding $i:P\to \hat P$ of a poset $P$ into its complete poset of downsets $\hat P$ is the free completion of the poset, in the sense that every poset morphism 
$P\to Q$, where $Q$ is complete, has a unique factorization through $i$ preserving all joins. However, $[-\infty,\infty]$ is not quite the free completion of $\mathbb{Q}$. Indeed, there is no factorization of $i:\mathbb Q\to \widehat{\mathbb Q}$ through the inclusion $\mathbb{Q}\to [-\infty,\infty]$ preserving joins, since the join of the set $\{1-1/n:n\in\mathbb N\}$ in $[-\infty,\infty]$ is $1$, while in $\widehat{\mathbb Q}$ it is $\{x:x<1\}$. 
The property that $[-\infty,\infty]$ does have, and more generally that characterizes Dedekind-MacNeille completions $j:P\to \bar P$ of posets, is that $[-\infty,\infty]$ is the universal complete poset admitting a map from $\mathbb{Q}$ which preserves joins which already exist. This is in contrast to $i:\mathbb Q\to \widehat{\mathbb Q}$, which destroys the join mentioned at the end of the last paragraph. 
Indeed free cocompletions destroy whatever colimits in a given enriched setting are not absolute-joints of finite sets in posets, splitting of idempotents in ordinary categories, biproducts in categories enriched over abelian groups, etc. So if $Q$ is some complete poset, then maps $\mathbb Q\to Q$ which preserve joins factor uniquely through $[-\infty,\infty]$. We could say that $[-\infty,\infty]$ is the free conservative cocompletion of $\mathbb{Q}$-the cocompletion that doesn't mess up anything we've already handled.
Arguably the Dedekind completion is a bit more abstruse, categorically, than the Cauchy completion. If one views $\mathbb{Q}$ as a category enriched over $[0,\infty]$, that is, as a generalized metric space, then the Cauchy completion is just the closure of $\mathbb{Q}$ in its category of enriched presheaves under absolute colimits-the absolute colimits in metric spaces being essentially limits of Cauchy sequences. For this reason, the closures of other kinds of enriched categories under absolute colimits are sometimes also called Cauchy completions. 
