# Example(s) of monoidal symmetric closed category with NNO without infinite coproducts?

The question is in the title, here is my motivation:

$\require{AMScd}$Let $(\mathcal C,\otimes,I)$ be a monoidal symmetric closed category. Then, the tensor product commutes with colimits, and if $\mathcal C$ has infinite coproducts, the object $N:=\displaystyle\coprod_{n\in\mathbb N}I$ has the following properties :

• it is a natural numbers object (NNO) in the sense of Lawvere, ie it has natural morphisms $0:I\to N$ and $S:N \to N$ such that any given $I\to X\to X$ uniquely defines a morphism $f:N\to X$ such that the following diagram commutes

$$\begin{CD} I @>0>> N @>S>> N\\ @| @VfVV @VfVV\\ I @>>> X @>>> X \end{CD}$$

• it is also the free monoid on $I$, ie the initial object in the category of monoids under $I$

If such a coproduct does not exist, a natural numbers object (if it exists) is the free monoid on $I$, but the converse is not necessarily true (or at least, I haven't been able to prove it). To better understand what might happen, It would help to see a workable example of such a category.

Of course for such an example to be interesting, I would like to have a category where an NNO or the free monoid exists.

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Trivial example: the category of finite sets with the cartesian product... – Zhen Lin Jun 26 '14 at 8:14
Actually, I would like a category where a Natural Number Object could exist. I have edited my question accordingly. – Cyrille Corpet Jun 26 '14 at 8:23
I don't get your definition of NNO. It looks like incomplete. – Fernando Muro Jun 26 '14 at 8:59
This definition allows to define sequences in $X$ by recursion: given a first "element" $1\to X$ and a "recursion law" $X\to X$, it defines a sequence $N\to N$. Nicer properties are given by the symmetric closed structure, for instance, addition is given as a map $N\otimes N\to N$ which is defined by adjunction by a map $N\to \operatorname{Hom}(N,N)$, itself defined by the usual recursive definition in $\operatorname{\mathbf{Set}}$. – Cyrille Corpet Jun 26 '14 at 9:09
Could you add "with NNO" to the title of your question? (A counterexample to the question in your title, but which I think has no NNO anyway, is the category of Banach spaces and continuous linear maps between them.) – Yemon Choi Jun 26 '14 at 12:33

The effective topos is an example of a (locally!) cartesian closed category with NNO where infinite coproducts do not exist. Indeed, the main feature of the NNO $N$ in the effective topos is that the endomorphisms of $N$ are precisely the computable functions. As such, $N$ cannot be (isomorphic to) $\coprod_{n \in \mathbb{N}} 1$: if that were the case, then there would be uncountably many endomorphisms of $N$. (Of course, $\coprod_{n \in \mathbb{N}} 1$ does not even exist in the effective topos; as you say, if it existed, it would have to be the NNO.)
Here is a closely related construction. Suppose we have a countable model $M$ of Zermelo set theory, or even just Mac Lane set theory. (For instance, by Skolem, we may take $M$ to be a countable elementary substructure of $V_{\omega + \omega}$.) Then the category of sets in $M$ is an elementary topos (so locally cartesian closed) with NNO, but its NNO cannot be $\coprod_{n \in \mathbb{N}} 1$, for the same reason.
Another example is a filterquotient topos. E.g. let $\mathcal{U}$ be a nonprincipal ultrafilter on $\mathbb{N}$ and consider the filterquotient of $Set/\mathbb{N}$ by $\mathcal{U}$. This is a world of "nonstandard analysis" which has a NNO (the image of $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$) but not every external infinite sequence of elements is an internal one, so the NNO is not $\coprod_{n\in\mathbb{N}} 1$.