Abject apologies for being a bit late to this party; had a bit of trouble getting a pumpkin (can’t be a Norwegian myth).

All of the suggested starting categories have a separator and this has serious consequences for Isbell conjugation. (To throw a link to the discussion over there, after working some of this out on paper I tried a few MathSciNet searches to see if anyone else had come up with it; several of the searches I tried used MSC codes to try to restrict the results to something useful. For some reason, the words “conjugation” and “dual” are fairly common in mathematics.)

So let $\mathcal{C}$ be a category with a separator, say $S$. Let $X$ be a presheaf on $\mathcal{C}$ (oh how I’d love to be able to do macros in iTeX); that is, as I’ve learnt, a contravariant functor from $\mathcal{C}$ to $Set$.

For objects $A, B$ in $\mathcal{C}$ we therefore get a set map

$\mathcal{C}(A,B) \to Set(X(B), X(A))$

functorial in both $A$ and $B$. We can turn this around slightly to get a set map

$X(B) \to Set(\mathcal{C}(A,B), X(A))$

still functorial in both $A$ and $B$. In terms of elements (gosh, did I really use that word?), this is

$x \mapsto \big( f \mapsto X(f)x \big)$

In particular, for a $\mathcal{C}$-morphism $g : B \to C$ we get a commutative diagram

$\begin{aligned}
X(C) & \overset{\quad\quad}{\to} && Set(\mathcal{C}(S,C), X(S)) \\
X(g) \downarrow & && \downarrow \\
X(B) & \overset{\quad\quad}{\to} && Set(\mathcal{C}(S,B), X(S))
\end{aligned}$

where the right-hand vertical arrow is the obvious one. Hence we can define a new functor $Y : \mathcal{C}^{op} \to Set$ by

$Y(B) = Im\big( X(B) \to Set(\mathcal{C}(S,B), X(B)) \big)$

There is an obvious natural transformation $X \to Y$ which consists of surjections.

Let us denote Isbell conjugation by primes. From the natural transformation $X \to Y$ we obtain a natural transformation $Y' \to X'$. I claim that this is a natural *isomorphism*.

Let us start with injectivity.

Let $B$ be an object of $\mathcal{C}$ and let $f \in X'(B)$. This is a natural transformation $X \to B^*$ so for an object $A$ of $\mathcal{C}$ we have a map of sets

$f_A : X(A) \to B^*(A) = \mathcal{C}(A,B).$
In particular, we get $f_S : X(S) \to \mathcal{C}(S,B)$.

Suppose that $f, g \in X'(B)$ are such that $f \ne g$. Then there is some object $A$ of $\mathcal{C}$ such that $f_A \ne g_A : X(A) \to \mathcal{C}(A,B)$. As these are set maps, there is some $x \in X(A)$ such that $f_A(x) \ne g_A(x) \in \mathcal{C}(A,B)$. These are now morphisms in $\mathcal{C}$. As $S$ is a separator, there is some morphism $a : S \to A$ such that $f_A(x)a \ne g_A(x)a$.
Now as $f$ and $g$ are natural transformations,

$f_A(x)a = \mathcal{C}(a, B) f_A(x) = f_S (X(a)x)$

and similarly for $g$. Hence the element $X(a)x \in X(S)$ distinguishes $f_S$ and $g_S$. Thus the map

$X'(B) \to Set(X(S), \mathcal{C}(S,B)), \quad f \mapsto f_S$

is injective.

This holds with $Y$ in place of $X$ and we have a commuting diagram.

$\begin{aligned}
X'(B) & \overset{\quad\quad}{\to} && Set(X(S), \mathcal{C}(S,B)) \\
\uparrow &&& \uparrow \\
Y'(B) & \overset{\quad\quad}{\to} && Set(Y(S), \mathcal{C}(S,B))
\end{aligned}$

with both horizontal maps being inclusions.

Now

$Y(S) = Im\big(X(S) \to Set(\mathcal{C}(S,S), X(S))\big)$

where the map is $x \mapsto (f \mapsto X(f)x)$. There is also a map

$Set(\mathcal{C}(S,S), X(S)) \to X(S)$

given by evaluation on the identity. The composition is the identity on $X(S)$ and thus $Y(S) \cong X(S)$. Moreover, this isomorphism is that appearing in the natural transformation $X \to Y$. Thus in the above diagram, the right-hand vertical map is an isomorphism. The left-hand map is thus injective.

Now let us consider surjectivity.

Let $f \in X'(B)$. This is a natural transformation $X \to B^*$. We need to show that it factors through $Y$. Let $A$ be an object in $\mathcal{C}$ and suppose that $x, y \in X(A)$ are such that $f_A(x) \ne f_A(y)$ in $\mathcal{C}(A,B)$. Then there is a morphism $a : S \to A$ such that $f_A(x)a \ne f_A(y)a$. Using again the fact that $f$ is a natural transformation, $f_A(x)a = f_S(X(a)x)$ and similarly for $y$. Thus $X(a)x \ne X(a)y$ in $X(S)$. Hence the maps $\mathcal{C}(S,A) \to X(S)$ defined by $x$ and $y$ differ. That is, the images of $x$ and $y$ in $Y(A)$ are distinct.

This shows that $f_A$ factors through $Y(A)$ and hence, as $X \to Y$ consists of surjections, we can build a natural transformation $f^Y : Y \to B^*$, such that under $Y' \to X'$, $f^Y$ maps to $f$. Thus $Y'(B) \to X'(B)$ is surjective.

We therefore conclude that the natural transformation $Y' \to X'$ is a natural isomorphism.

We have

$\begin{aligned}
Y(A) &\subseteq Set(\mathcal{C}(S,A), X(S)) \\
Y'(A) &\subseteq Set(X(S), \mathcal{C}(S,B))
\end{aligned}$

we also have a pairing compatible with these inclusions; that is (on objects)

$Y(A) \times Y'(B) \to \mathcal{C}(A,B) \subseteq Set\big(\mathcal{C}(S,A), \mathcal{C}(S,B)\big)$

(inclusion as $S$ is a separator).

From this it is clear that $Y''$ is given by

$Y''(A) = \big\{ c : \mathcal{C}(S,A) \to X(A) \mid f c \in \mathcal{C}(A,B) \forall B \in \mathcal{C}, f \in Y'(B)\big\}$

Lo and behold! A Frölicher space! Or rather, A Frölicher $\mathcal{C}$-object.

Thus stability under Isbell conjugation is a very strong condition indeed; it forces *quasi-representability* and also the saturation condition. This suggests to me that there is a *qualitative* difference between quasi-representability and non-quasi-representable. As I conjectured, there is a functor from all presheaves to quasi-representable ones, and of course then one can (may I say “should”?) saturate with respect to duality. So every presheaf has an underlying Frölicher object, but this assignment is by no means injective. In fact, contrary to what I thought, the fibres of this functor do have some interesting stuff in them.

For example, if we take the simplified version of Urs’ favourite functor; namely 1-forms, then the associated Frölicher space is the single point (since $\Omega^1(pt) = \{0\}$).

It turns out, therefore, that Urs and I are looking at *orthogonal* concepts!

One possible area for further study is to see what happens if you vary the separator. It doesn’t have to be a singleton point in our categories of interest. For example, if we enlarge our category of “known smooth stuff” to include $\mathbb{R}^{\infty}$, the direct limit of the $\mathbb{R}^n$s, and $X = \Omega^1$ to be 1-forms, then by taking $\mathbb{R}^{\infty}$ as the separator we find that

$\Omega^1(M) \to Set\big(C^\infty(\mathbb{R}^{\infty},M), \Omega^1(\mathbb{R}^\infty)\big)$

is injective, and so $\Omega^1$ *is* quasi-representable. Of course, it is no longer quasi-representable in the strictest sense because we do not have an injection

$\Omega^1(M) \subseteq Set(\vert M \vert, X)$

for any set $X$. We only get this type of quasi-representability if our separator is a singleton point.

So perhaps we should define **quasi-$S$-representable** where $S$ is a separator in the category $\mathcal{C}$, by which we mean that the natural transformation which on objects is

$X(B) \to Set\big(\mathcal{C}(S,B), X(S)\big)$

consists of injections.

However, varying the singleton won’t change the fact that Isbell conjugation (now thinking of one of our smoothish categories) results in *strictly* quasi-representable presheaves since the argument above made no assumptions on the separator $S$ so it certainly does work for $S$ the singleton point.

Right, it’s lunchtime so I’m stopping here.

Andrew

## Re: Space and Quantity

I have restructured a bit and added a section (number 2 at the moment) on the general abstract nonsense of

space and quantity. I followed Lawvere’s Taking categories seriously, but varied slightly. Please check.(I haven’t said anything about the dotted arrows that Lawvere has on p. 17. I guess these are supposed to allude to the kind of “saturation” or “completion” operation which we keep talking about?)

I have added a statement, right now proposition 5, which says that for all $C^\infty DGCAs$ $A$ and $g$ any $L_\infty$-algebra we have

$CE(g)\otimes_\infty A \simeq CE(g) \otimes A \,.$

I think this is essentially obvious, using the fact that the GCA underlying $CE(g)$ is freely generated in positive degree. But please check.