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This is Exercise I.2 of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]."

The Question:

Prove that $\mathbf{FinSets}^{\mathbf{N}}$ has no subobject classifier.

Here $\mathbf{FinSets}$ is the category of objects all finite sets and arrows all functions between them. We denote by $\mathbf{N}$ the linearly ordered set of natural numbers.

A definition of a subobject classifier is given on page 32, ibid.

Definition: In a category $\mathbf{C}$ with finite limits, a subobject classifier is a monic, ${\rm true}:1\to\Omega$, such that to every monic $S\rightarrowtail X$ in $\mathbf{C}$ there is a unique arrow $\phi$ which, with the given monic, forms a pullback square

$$\begin{array}{ccc} S & \to & 1 \\ \downarrow & \, & \downarrow {\rm true}\\ X & \stackrel{\dashrightarrow}{\phi} & \Omega. \end{array}$$

Thoughts:

I'm not sure what the terminal object of $\mathbf{FinSets}^{\mathbf{N}}$ is, if it exists at all. My guess is that it's the functor $1: \Bbb N\to \{\ast\}, n\mapsto \ast$ for the singleton set $\{\ast\}$ up to isomorphism but my suspicion is that this guess is way off.

My idea so far is to take some monic $S\stackrel{f}{\rightarrowtail}X$ in the category in question and show, somehow, that there is no such arrow as ${\rm true}: 1\to\Omega$ satisfying the definition. I don't know yet how to execute this idea.

Further Context:

I have recently finished a light reading of Goldblatt's book, "Topoi: A Categorial Analysis of Logic". I have been interested in topoi for a good few years now. (See some of my very first questions on this site.)

I think, then, that I ought to be able to solve this myself. I'm keen to try out other questions, though, and this one is taking me too long.

Please help :)

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There's a standard trick using the Yoneda lemma for computing what universal objects in (restricted) functor categories must be if they exist. In the case of the subobject classifier, this is explained on p. 37 of Sheaves in Geometry and Logic.

Specifically, suppose that $\Omega\colon \mathbf{N}\to \mathbf{FinSets}$ is a subobject classifier in $\mathbf{FinSets}^\mathbf{N}$. Let's try to find out what finite set $\Omega(0)$ is. Since $\mathbf{FinSets}^\mathbf{N}$ is a full subcategory of $\mathbf{Sets}^\mathbf{N}$, by Yoneda we have $$\Omega(0) \cong \text{Hom}_{\mathbf{Sets}^\mathbf{N}}(h^0,\Omega) = \text{Hom}_{\mathbf{FinSets}^\mathbf{N}}(h^0,\Omega) \cong \text{Sub}(h^0),$$ where $h^0$ is the functor $\text{Hom}_{\mathbf{N}}(0,-)$. But now $h^0$ has infinitely many subobjects, but $\Omega(0)$ is a finite set, and this is a contradiction.

To see that $h^0$ has infinitely many subobjects, just note that since $0$ is the initial object in $\mathbf{N}$, $h^0(n)$ is a singleton $\{*\}$ for all $n$. Incidentally, this makes $h^0$ isomorphic to the terminal object $1$ - your guess about the identity of terminal object is correct.

Now for each natural number $n$ (or $n = \infty$), there is a distinct subobject of $h^0$, given by $$m\mapsto \begin{cases} \varnothing & \text{if }m<n\\ \{*\} & \text{if }m\geq n.\end{cases}$$

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You are actually right that that the terminal object of your category is the constant functor at $\{βˆ—\}$: what is a morphism to the constant functor at $\{βˆ—\}$? It is a general fact about functor categories: limits can be computed pointwise (if the limit exists pointwise, then it exists globally as a functor; but the converse is not true).

But actually, you can proceed like this without using which object is $1$: try to find an object $X$ such that for every possible $𝛺$, the number of arrows to $𝛺$ is too small to classify the subobjects of $X$. Write things down: fix $X = (X_1 β†’ X_2 β†’ X_3 β†’β€ˆβ‹―)$ and $𝛺 = (𝛺_1 β†’ 𝛺_2 β†’ β‹―)$. What is an arrow from $X$ to $𝛺$? There is potentially lots of them. How can you make this quantity smaller without putting too much restriction on $X$?

Here is a further hint: think of the process of building an arrows $X→𝛺$ as follows. First choose an arrow $X_1→𝛺_1$, then an arrow $X_2→𝛺_2$ making the diagram commute, then $X_3→𝛺_3$, etc. How can you make it so that at each step, there is exactly one choice?

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Regarding the Terminal Object:

You are correct that $\newcommand\1{\mathbf{1}}$the terminal object is the constant functor $\1:n\mapsto \newcommand\set[1]{\left\{{#1}\right\}}\set{*}$.

This is more generally true in any functor category $[C,D]$. If $D$ has a terminal object, $t$, then the constant functor $\Delta_t : c\mapsto t$ is a terminal object in $[C,D]$, since for any functor $F:C\to D$, there is a unique map $\alpha_c : Fc \to \Delta_t c= t$, since $t$ is a terminal object in $D$. You can check that $t$ being terminal also causes $\alpha$ to be natural.

Note:

The terminal object will prove to be irrelevant. It suffices to know that in our case, $\Omega$ (if it existed) would represent the subobject functor.

Subobjects in $[\mathbb{N},\mathbf{FinSet}]$:

Now we need to figure out what subobjects look like. First observe that if a natural transformation $\alpha :F \to G$ is such that $\alpha_n$ is injective for all $n$, then it is a monomorphism. This is because if $\alpha\beta= \alpha\gamma$, then $\alpha_n\beta_n = \alpha_n\gamma_n$ for all $n$, and then since $\alpha_n$ is injective, $\beta_n=\gamma_n$ for all $n$. Therefore $\beta=\gamma$. I will call this property being a pointwise monomorphism.

Note:

This generalizes to general functor categories, pretty immediately.

Aside, Pointwise Monomorphisms are equivalent to Monomorphisms in $[\mathbb{N},\mathbf{FinSet}]$:

Consider the functor $(-)_n : [\newcommand\FinSet{\mathbf{FinSet}}\newcommand\NN{\mathbb{N}}\NN, \FinSet] \to \FinSet$ that sends $F$ to $Fn$, for $n\in\NN$. This functor has a left adjoint. For a set $A$, we define the functor $A_{\ge n}$, which sends $m$ to $\varnothing$ if $m < n$, and $m$ to $A$ if $m\ge n$. The vertical maps are the obvious ones (inclusion of $\varnothing$ into $A$ or the identities). You can verify that $$[\NN,\FinSet](A_{\ge n},F) \simeq \FinSet(A,(F)_n).$$

Thus the functor $(-)_n$ is a right adjoint, and therefore preserves monomorphisms. Hence monomorphisms are pointwise monomorphisms in our category.

Nonexistence of a Subobject Classifier:

Suppose a subobject classifier $\1\to \Omega$ existed. Let $\newcommand\Sub{\operatorname{Sub}}\Sub:[\NN,\FinSet]^{\text{op}}\to\mathbf{Set}$ be the subobject functor. Then $$[\NN,\FinSet](A_{\ge 0},\Omega) \simeq \FinSet(A,(\Omega)_0) \simeq \Sub(A_{\ge 0}).$$ Now the subobjects of $A_{\ge 0}$ are increasing chains of subsets of $A$ (regarded as functors in the obvious way). Let $\omega = \#(\Omega)_0$. Then if $A$ has cardinality $n$, $\#\FinSet(A,(\Omega)_0) = \omega^n$. I.e., the number of mappings have exponential growth in $n$. How about subobjects of $A_{\ge 0}$? Well a maximal chain of subsets of $A$ is equivalent to a permutation of $A$, so there are at least $n!$ subobjects of $A_{\ge 0}$. Thus the number of subobjects of $A_{\ge 0}$ grows faster than exponentially in the size of $A$. This is a contradiction.

Note:

We didn't actually need that monomorphisms were monomorphisms pointwise, which is why I called it an aside, but we did use that $(-)_0$ was right adjoint to $-_{\ge 0}$.

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