Category of comagmas in $\mathbf{Set}$ and other topoi A comagma, of course, is just the dual of a magma. Hence, a comagma in a category $C$ with finite coproducts is an object $X$ of $C$ together with a "cooperation" $X \to X \coprod X$.
For the case of $\mathbf{Set}$, or more generally, any extensive category, the "doubling" functor $X \mapsto X \coprod X$ preserves connected limits.
Hence, the following question:

Is the category of comagmas in $\mathbf{Set}$ or any other topos with a natural numbers object itself a topos?

 A: Here's a sketch of why I think the comagmas in Set form a topos. Let $S$ be the following category: An object is an infinite sequence of $0$'s and $1$'s. A morphism from one such sequence $s$ to another $t$ is $(s,n,t)$ where $n$ is a natural number such that deleting the first $n$ terms of $s$ produces $t$. Composition is done by adding the $n$'s; identity morphisms have $n=0$. All morphisms are obtainable by composition from the morphisms with $n=1$. 
Given any comagma $(X,q)$, each element $x\in X$ gives rise to a sequence $d(x)$, an object of $S$, as follows. The comagma structure $q$ of $X$ sends $x$ to an element $q(x)$ in one of the two summands of $X\sqcup X$, which I'll call the $0$th and $1$th summands. If $q(x)$ is the element $y$ of $X$ in the $i$th summand (i.e., the image of $y\in X$ under the $i$th coproduct injection $X\to X\sqcup X$), then the first term in $d(x)$ is $i$ and the rest of $d(x)$ is $d(y)$.  In other words, $d(x)$ records which of the two summands one lands in after applying $q$ to $x$ repeatedly.  
For each object $s$ of $S$, let $X_s=d^{-1}(\{s\})$. If $x\in X_s$ and if $t$ is the result of deleting the first term of $s$, then $X_t$ contains $q(x)$ (more precisely, the element of $X$ that you get from $q(x)\in X\sqcup X$ by forgetting which summand it's in). 
In this way we get a functor $S\to$ Sets, sending $s$ to $X_s$ and $(s,1,t)$ to $q$ (or the more precise version of $q$ above). I claim (and I won't have time to prove it here, which is why this is only a sketch) that (1) this is really a functor, hence an object $\tilde X$ of Set${}^S$, (2) that any homomorphism $X\to Y$ of comagmas induces a natural transformation $\tilde X\to\tilde Y$, in a coherent way, so that we have a functor from the category of comagmas to Set${}^S$, and (3) that this functor is an equivalence of categories.
If all this is correct, then of course it follows that the category of comagmas is a topos, the topos of presheaves on the dual of $S$.
Furthermore, I don't see that I've used very much about the category of sets here (or in the omitted stuff that isn't here), so I think the same construction will show that the comagmas in any topos $\mathcal E$ with natural numbers object form a topos, a topos of internal presheaves over $\mathcal E$'s version of $S^{\text{op}}$.
