3
$\begingroup$

This is a follow up to this question.

For a smooth manifold $M$ let $\mathbb{R}^M$ be the ring of smooth maps from $M$ to $\mathbb{R}$. Now let $C$ denote the category of all such rings, where arrows are the ring homomorphisms which corresponds to the smooth maps. Let $\mathrm{Set}^C$ denote the topos of presheafs of $C^{\mathrm{op}}$.

I am interested in knowing the subobjects of the terminal presheaf in $\mathrm{Set}^C$, as well as the subobject classifier of $\mathrm{Set}^C$.

As is explained in this answer. The subobjects of the terminal presheaf corresponds to the presheafs $F$ such that $F(c)$ is either singleton or empty for all object $c$.

For each manifold $M$, there exists a unique smooth map from $\mathrm{pt}$ to $M$ where $\mathrm{pt}$ is the $0$-dimensional manifold. Therefore for each $\mathbb{R}^N$ ($N$ is some manifold), there exists unique homomorphism from $\mathbb{R}^N$ to $\mathbb{R}^{\mathrm{pt}} \cong \mathbb{R}$. So $\mathrm{Hom}(\mathbb{R}^M,\mathbb{R}^N)$ is a singleton for all $M$ when $N = \mathrm{pt}$. Are there any other $N$ such that this is true?

Moreover, are there any $N$ such that $\text{Hom}(\mathbb{R}^M,\mathbb{R}^N)$ is singleton for some $M$ and empty for some others? I think this is not true, which then make me think that $\mathrm{Set}^{C}$ is a boolean. Is this correct?

Thanks in advance!!

$\endgroup$
2
  • 2
    $\begingroup$ Presheaf toposes are almost never boolean. (This one isn't.) $\endgroup$
    – Zhen Lin
    Commented Jan 25, 2023 at 4:22
  • 1
    $\begingroup$ The category $C$ that you defined is in fact equivalent to the category of smooth manifolds. $\endgroup$
    – Dmitri P.
    Commented Jan 25, 2023 at 6:58

1 Answer 1

2
$\begingroup$

I'm assuming that $C^\mathrm{op}$ is equivalent to the category of smooth manifolds here, see the comment by Dmitri P.

The singleton $\mathrm{pt}$ has the property that $\mathrm{Hom}(N,\mathrm{pt})$ is a singleton for all smooth manifolds $N$. In other words, it is a terminal object in the category of smooth manifolds. Terminal objects are always unique (up to unique isomorphism), so there are no other smooth manifolds with this same property.

The image of $\mathrm{pt}$ under the Yoneda embedding is the terminal object in $\mathrm{Set}^C$.

Now let $F$ be a subobject of the terminal object (also called a subterminal object). Then $F(N)$ is either the empty set or a singleton, for each smooth manifold $N$. There is a unique map $N \to \mathrm{pt}$, and this induces a morphism $F(\mathrm{pt}) \to F(N)$. As a result, if $F(\mathrm{pt})$ is nonempty, then $F(N)$ is nonempty as well for $N$. But in this case we get that $F(N)$ is a singleton for each $N$. Now suppose that $F(\mathrm{pt})$ is empty. If a smooth manifold $N$ has a point $x : \mathrm{pt} \to N$, then there is an induced morphism $F(N) \to F(\mathrm{pt})$, and because $F(\mathrm{pt})$ is empty, $F(N)$ must be empty as well. However, there are two possibilities for $F(\varnothing)$: the set $F(\varnothing)$ can either be empty or a singleton.

In conclusion, there are three possibilities for a subterminal object $F$. First of all, there is the terminal object itself, with $F(N)$ a singleton for each $N$. Then there is also the initial object, with $F(N)$ the empty set for each $N$. Finally, there is the presheaf $F$ with $F(N)$ the empty set whenever $N$ has a point, and with $F(\varnothing)$ a singleton.

The subobject classifier is by definition a presheaf $\Omega$ such that $\Omega(N)$ is the set of subobjects of the image of $N$ under the Yoneda embedding. By the above, we already know that $\Omega(\mathrm{pt})$ is a set with three elements. The sets $\Omega(N)$ for general $N$ will be more difficult to describe explicitly. For example, different open submanifolds of $N$ will determine different elements of $\Omega(N)$, so typically $\Omega(N)$ will be infinite.

$\endgroup$
1
  • 2
    $\begingroup$ More concisely: the terminal presehaf always consists of singletons, a presheaf $F$ is subterminal if it consists of singletons or empty sets with the property that if $F(Y)$ is a singleton, then the existence of a morphism $X\to Y$ implies $F(X)$ is a singleton. The fact that in manfiolds a morphism $X\to Y$ fails to exist only when $Y$ is empty and $X$ is not implies that $F(X)$ is identically a singleton or empty for all non-empty $X$, and in the latter case $F(\emptyset)$ is also empty. $\endgroup$ Commented Jan 26, 2023 at 18:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .