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1. • CommentRowNumber1.
   • CommentAuthorTodd_Trimble
   • CommentTimeDec 21st 2014
   • (edited Dec 21st 2014)
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexHere's something I'd never noticed before yesterday: in the category of posets, letting $Down(P)$ denote the internal hom $[P^{op}, \mathbf{2}]$, there is at most one retraction $\pi$ of the yoneda = principal down-set embedding $y: P \to Down(P)$, and this occurs precisely when $\pi \dashv y$ (so that $P$ is a sup-lattice). A "lowbrow" proof might go like this: if $D \subseteq P$ is a down-set, then for all $d \in D$ we have $y(d) \subseteq D$, whence $d = \pi y(d) \leq \pi(D)$, showing that $\pi(D)$ is an upper bound of $D$. On the other hand, if $e$ is any upper bound of $D$, then $D \subseteq y(e)$, whence $\pi(D) \leq \pi y(e) = e$, so that $\pi(D)$ is a least upper bound of $D$. I don't know why exactly, but I found this slightly disconcerting. Offhand, I would have expected *many* retractions are possible. But there's at most one! Slightly more generally: suppose $V$ is a commutative quantale (I'm gearing up to general enriched categories) which we can think of as a small [[cosmos]], and suppose $P$ is a $V$-enriched category. If $f: P^{op} \to V$ is a $V$-functor, then we can use the enriched so-called "co-yoneda" lemma to write $$f = \int^p f(p) \cdot y(p)$$ and now if we suppose $\pi: V^{P^{op}} \to P$ is a $V$-functor retraction of $y$, then we may put $e = y \circ \pi: V^{P^{op}} \to V^{P^{op}}$. The enrichment of $e$ yields a canonical transformation $$f = \int^p f(p) \cdot e y(p) \to e(f)$$ (where the first equation uses $e y = y \pi y = y$) which gives the unit $1 \to y\pi$ of an adjunction $\pi \dashv y$, with counit the retraction isomorphism $\pi y \stackrel{\sim}{\to} 1_P$. So again, there's only one possible retraction. For a while I thought this type of calculation might generalize to a general cosmos $V$. If there is $\pi: V^{C^{op}} \to C$ together with a $V$-natural isomorphism $\pi y_C \to 1_C$, one can indeed manufacture a $V$-natural candidate for a unit $1 \to y_C \pi$ along the above lines, but I wasn't able to see the triangular equations (which come for free in posetal cases like the above). (Not even for the case $V = Set$.) Either this is because I'm blind or stupid here, or in fact in general there is no such adjunction. Which is it? So I'm putting this question to readers here: is there an example of a locally small category $C$ and a functor $F: Set^{C^{op}} \to C$ and an invertible transformation $F y_C \to 1_C$ but where $F$ is *not* left adjoint to $y_C$?

   Here’s something I’d never noticed before yesterday: in the category of posets, letting $Down(P)$ denote the internal hom $[P^{op}, \mathbf{2}]$, there is at most one retraction $\pi$ of the yoneda = principal down-set embedding $y: P \to Down(P)$, and this occurs precisely when $\pi \dashv y$ (so that $P$ is a sup-lattice).

   A “lowbrow” proof might go like this: if $D \subseteq P$ is a down-set, then for all $d \in D$ we have $y(d) \subseteq D$, whence $d = \pi y(d) \leq \pi(D)$, showing that $\pi(D)$ is an upper bound of $D$. On the other hand, if $e$ is any upper bound of $D$, then $D \subseteq y(e)$, whence $\pi(D) \leq \pi y(e) = e$, so that $\pi(D)$ is a least upper bound of $D$.

   I don’t know why exactly, but I found this slightly disconcerting. Offhand, I would have expected many retractions are possible. But there’s at most one!

   Slightly more generally: suppose $V$ is a commutative quantale (I’m gearing up to general enriched categories) which we can think of as a small cosmos, and suppose $P$ is a $V$-enriched category. If $f: P^{op} \to V$ is a $V$-functor, then we can use the enriched so-called “co-yoneda” lemma to write

   $$f = \int^p f(p) \cdot y(p)$$

   and now if we suppose $\pi: V^{P^{op}} \to P$ is a $V$-functor retraction of $y$, then we may put $e = y \circ \pi: V^{P^{op}} \to V^{P^{op}}$. The enrichment of $e$ yields a canonical transformation

   $$f = \int^p f(p) \cdot e y(p) \to e(f)$$

   (where the first equation uses $e y = y \pi y = y$) which gives the unit $1 \to y\pi$ of an adjunction $\pi \dashv y$, with counit the retraction isomorphism $\pi y \stackrel{\sim}{\to} 1_P$. So again, there’s only one possible retraction.

   For a while I thought this type of calculation might generalize to a general cosmos $V$. If there is $\pi: V^{C^{op}} \to C$ together with a $V$-natural isomorphism $\pi y_C \to 1_C$, one can indeed manufacture a $V$-natural candidate for a unit $1 \to y_C \pi$ along the above lines, but I wasn’t able to see the triangular equations (which come for free in posetal cases like the above). (Not even for the case $V = Set$.) Either this is because I’m blind or stupid here, or in fact in general there is no such adjunction. Which is it?

   So I’m putting this question to readers here: is there an example of a locally small category $C$ and a functor $F: Set^{C^{op}} \to C$ and an invertible transformation $F y_C \to 1_C$ but where $F$ is not left adjoint to $y_C$?

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeDec 22nd 2014
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexInteresting question! At first I thought I could also generalize your calculation, but now I think I was wrong. I do think I can prove that if such an $F$ exists, then $C$ is "weakly total", in that it has weak colimits (existence but not uniqueness of factorizations) of all the diagrams that a total category would have colimits of, and $F$ assigns those weak colimits. Conversely, if $C$ is weakly total and we can choose those weak colimits functorially in a way that assigns the actual colimit of each representable, then that functorial choice ought to be such an $F$. Can we think of any weakly total category that is not total?

   Interesting question! At first I thought I could also generalize your calculation, but now I think I was wrong.

   I do think I can prove that if such an $F$ exists, then $C$ is “weakly total”, in that it has weak colimits (existence but not uniqueness of factorizations) of all the diagrams that a total category would have colimits of, and $F$ assigns those weak colimits. Conversely, if $C$ is weakly total and we can choose those weak colimits functorially in a way that assigns the actual colimit of each representable, then that functorial choice ought to be such an $F$. Can we think of any weakly total category that is not total?