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An algebraic theory is a small category $\cal T$ with finite products.An algebra for the theory $\cal T$ is a functor $A:\cal T\to Set$ preserving finite products. We denote by $Alg\ \cal T$ the category of algebras of $\cal T$. The homomorphisms are natural transformations. Thus $Alg\ \cal T$ is a full subcategory of the functor category $\cal Set^T$.

Now an algebraic theory $\cal N$ for $\cal Set$ can be described as the full subcategory of $\cal Set^{op}$ whose objects are the natural numbers $n=\{0,1,...,n-1\}$. In fact, since $n=1\times...\times 1$ in $\cal Set^{op}$ is determined up to isomorphism by the set $A1$.

More precisely, we have an equivalence functor $$E:Alg{\cal\ N\to Set},\ \text A\mapsto A1.$$

My question is what do the morphisms of $Alg\cal\ N$ look like (i.e. how can they be described and understood) and how does $E$ act on them?

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The subcategory of $\mathbf{Set}^{op}$ you describe is the (opposite of the) skeleton of $\mathbf{FinSet}$, the category of finite sets and functions, and is sometimes called $\mathbf{FinOrd}$. $\mathbf{FinOrd}$ is the category with finite coproducts freely generated from a single object, thus $\mathbf{FinOrd}^{op}$ is the category with finite products freely generated from a single object which is its significance here.

As you've stated, the arrows of $\mathsf{Alg}\,\mathbf{FinOrd}^{op}$ are the natural transformations. I'll write $[n]$ for the $n$-fold product of the generating object in $\mathbf{FinOrd}^{op}$ which is thus called $[1]$ and the terminal object, i.e. the nullary product, is thus $[0]$. Since functors $F,G\in\mathsf{Ob}(\mathsf{Alg}\,\mathbf{FinOrd}^{op})$ preserve finite products, they send projections to projections, and tupling to tupling. In particular, given a natural transformation $\tau:F\to G$, we have the following naturality square:$$\require{AMScd}\begin{CD}F[1]\times F[1]@>\tau_{[2]}>> G[1]\times G[1]\\ @V\pi_1VV @VV\pi_1V \\ F[1] @>>\tau_{[1]}> G[1] \end{CD}$$ where I've implicitly used $F[2]=F([1]\times[1])\cong F[1]\times F[1]$ and $F\pi_1 = \pi_1$ and similarly for $G$ and $\pi_2$. This states that $\tau_{[2]}=\langle \tau_{[1]},\tau_{[1]}\rangle$, i.e. the tupling of $\tau_{[1]}$ with itself. It should be fairly obvious that similar reasoning leads to every component of $\tau$ being just tuplings of $\tau_{[1]}$. Unsurprisingly, the action of $E$ on $\tau$ is simply to take the $\tau_{[1]}$ component. It should also be clear how to invert $E$ too.

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  • $\begingroup$ Well, let $H$ be the inversion of $E$. How do I argue that $H\circ E \cong 1_{\mathsf{Alg}\,\mathbf{FinOrd}^{op}}$? $\endgroup$ – user122424 Feb 28 '18 at 16:10

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