# Lawvere theories: an equivalence.

I'm having trouble understanding Lawvere theories (as defined below).

Definition: A Lawvere Theory is a category $\mathcal{L}$ with finite products and with a distinguished object $A$ such that every object of $\mathcal{L}$ is (isomorphic to) a finite power of $A$; that is, for any $X\in\operatorname{Ob}(\mathcal{L})$, there is an $n\in\mathbb{N}$ with $X\cong A^{n}$. The object $A$ is called the fundamental object of $\mathcal{L}$. An arrow $\omega :A^{n}\to A$ is called an $n$-ary operation (and, in particular, arrows of the type $A^{0}=1\to A$ are called constants).

Here's an exercise in Turi's Category Theory Lecture Notes:

Let $\mathbb{N}^{\text{op}}$ be the opposite category of natural numbers and all functions. Show that Lawvere theories are equivalent to product preserving functors $$\mathbb{N}^{\text{op}}\to\mathbf{C}$$ that are bijective on objects.

The problem: I'm not sure what to do. The $\mathbf{C}$ doesn't help as it's undefined.

My attempt: Let $\mathcal{L}:\mathbb{N}^{\text{op}}\to\mathbf{C}$ be a product preserving functor bijective on objects. Then for any $(n_{i})_{I}\in\mathbb{N}$ for any set $I$, if $\prod_{i\in I}{n_{i}}$ exits, then $\mathcal{L}\left(\prod_{i\in I}{n_{i}}\right) = \prod_{i\in I}{\mathcal{L}(n_{i})}$ and for all $m, n\in\mathbb{N}$, $c\in Ob(\mathbf{C})$,

• $\mathcal{L}(n)=\mathcal{L}(m)\Rightarrow n=m$
• there exists $m_{c}\in\mathbb{N}$ with $\mathcal{L}(m_{c})=c$.

I then set up the commutative diagram(s) for the product with an arbitrary $(f_{i})_{I}$ such that $f_{i}: k\to n_{i}$ in $\mathbb{N}^{\text{op}}$ and took $\mathcal{L}$ of everything.

Then I got stuck.

I want to force this $\mathcal{L}$ to be a Lawvere theory. Once I've done that, I'll take a Lawvere theory and try to go in the other direction.

Thank you :)

Second attempt (based on the comments): The trick is to consider what happens to $1\in Ob(\mathbb{N}^{\text{op}})=\mathbb{N}$. Note that the product in $\mathbb{N}^{\text{op}}$ is the coproduct in $\mathbb{N}$, so is simply addition.

Suppose $L:\mathbb{N}^{\text{op}}\to\mathbb{C}$ is a product preserving functor bijective on objects. Then for all $c\in Ob(\mathbf{C})$, there exists a unique $m_{c}\in Ob(\mathbb{N}^{\text{op}})=\mathbb{N}$ with $L(m_{c})=c$ and for any $m, n\in\mathbb{N}, L(m)=L(n)\Rightarrow m=n$.

Let $L(1)=A$ and note that if a product $\sum_{i\in I}{a_{i}}$ exists in $\mathbb{N}^{\text{op}}$ for $a_{i}\in Ob(\mathbb{N}^{\text{op}})=\mathbb{N}$, it is in $\mathbb{N}$ and so $I$ would be a finite set. Hence finite products must exist in $\mathbf{C}$. Now for $X\in\mathbf{C}$, $X=L(m_{X})$ for some $m_{X}\in\mathbb{N}$ so $X=L(m_{X})=L(\sum_{j=1}^{m_{X}}{1})=\prod_{j=1}^{m_{X}}{L(1)}=A^{m_{X}}$. Thus $X\cong A^{m_{X}}$. Therefore, the pair $\langle L, \mathbf{C}\rangle$ is a Lawvere theory.

Ideas for the converse: Let $L:\mathbb{N}^{\text{op}}\to\mathcal{L}$ such that $L(1)=A$ and . . .

• $L(u)=U\cong V=L(v)$ iff $L(u)=L(v),$
• $Y\cong A^{m}=L(m_{A^{m}})$ implies $L(m_{A^{m}})=Y$, or
• $L(m)=Y$ if (and only if) $Y\cong A^{m}$.

I've explored these ideas and they don't seem fruitful. I would like an explicit proof now please, $\color{red}{\large\text{not just hints}}$. This is really bugging me.

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The category $\mathcal{C}$ is a Lawvere theory, if you have a bijective-on-objects product-preserving functor $\mathbb{N}^\mathrm{op} \to \mathcal{C}$. –  Zhen Lin Nov 29 '13 at 17:29
@Zhen Lin: Thank you. Given the definition above, though, I'm afraid that's little more than an assertion to me at the moment; I want to be able to show it :/ –  Alice Nov 29 '13 at 17:34
It says nothing of the sort. It says "equivalent". More precisely there is an equivalence between the category of Lawvere theories and the category of bijective-on-objects product-preserving functors with domain $\mathbb{N}^\mathrm{op}$, if you choose the appropriate notion of morphism on both sides. –  Zhen Lin Nov 29 '13 at 17:52
Yes, you can do that. –  Zhen Lin Nov 29 '13 at 17:57
The product in $\mathbb{N}^\mathrm{op}$ is the coproduct in $\mathbb{N}$, which is addition. –  Zhen Lin Dec 1 '13 at 22:41

Let $\mathbf C$ be a Lawvere theory: i.e. a category with finite products such that there's an object $A \in \mathbf C$ for which for every other $X \in \mathbf C$ there's a $n \in \mathbb N$ such that $A^n\cong X$.

Let's consider $\mathbb N$ the category of natural numbers and all function between them.

Clearly we have a function between the objects of $\mathbb N$ and the objects of $\mathbf C$ defined as $$\mathcal L\colon \mathbb N \to \mathbf C,$$ with $\mathcal L(n) = A^n$ for $n \in \mathbb N \setminus\{0\}$ and $\mathcal L(0)=\bullet$ the terminal object of $\mathbf C$.

Let $f \colon n \to m$ be a morphism in $\mathbb N$, or else a function between the sets $\{0,\dots,n-1\}$ and $\{0,\dots,m-1\}$.

For every $i \in \{0,\dots,m-1\}$ we can consider the family of morphism $\langle \pi_i^j\rangle_{j=1,\dots,n}$ where $\pi^i_j=1_A$ if $i \in f^{-1}(\{j\})$ other wise $\pi^i_j \colon A \to \bullet$ is the unique map in the terminal object $\bullet \in \mathbf C$.

These morphisms give us a morphism $$\pi^i \colon A \to A^{f^{-1}(\{i\})}$$ for universal property of products and then we get the product morphism $$\mathcal L(f) = \prod_{i=1}^n \pi^i \colon A^n=\mathcal L(n) \to A^m=\mathcal L(m).$$

Clearly if $f=1_n$ for a $n$ then for every $i \in n$ we have $1_n^{-1}(\{i\})=\{i\}$ and so the $\pi^i=1_A$ for every $i$ and so $\mathcal L(1_n)=1_{A^n}$.

Doing the calculations you can prove also that $\mathcal L(g \circ f)=\mathcal L(f) \circ \mathcal L(g)$, for every pair $f \colon n \to m$ and $g \colon m \to k$, of course the calculation are a little complicated (I would rather not to write here).

This functor is clearly product preserving, indeed for every $n \in \mathbb N$ we have that $\mathcal L(n)=A^n$ which is the n-fold product, and every projection of $\mathbb N^\text{op}$ i.e. a map $p \colon 1 \to n$ in $\mathbb N$ we have that

$$\mathcal L(p) = \prod_{i=1}^n \pi^i \colon A^n \to A,$$ where every $\pi^i \colon A \to \bullet$ for $i \not \in \text{Im }p(0)$ and $\pi^i = 1_A$ for $i \in \text{Im }p(0)$. So $\mathcal L(p)$ is the $p(0)$-th projection of the product $A^n$, of course now one should verify that these data verify the universal property which involve again some long calculations.

Forgive the lack of some details, but I think that adding them wouldn't make the answer more clear.

Hope this helps.

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Thank you so much for your patience in writing that out for me, @Giorgio Mossa! I'm confident that I can take it from here. [I first need to check the existence of $\bullet$.] I hope you understand if I accept once I've got the damn thing! –  Alice Dec 4 '13 at 11:58
@Shaun sure of course. Btw $\bullet$ the terminal object is also the empty product so it must exists in every category with finite products. :) –  Giorgio Mossa Dec 4 '13 at 12:08
Again, thank you. I'm not sure I understand how $\mathcal{L}$ is bijective on objects though. What if there exists an $X$ such that $A^{n}\neq X\cong A^{n}$? [The same question, I suppose, covers how we can overlook $A^{n}\times\bullet\cong A^{n}$.] –  Alice Dec 4 '13 at 19:23
Oh, I see; it's a bit of a misnomer. –  Alice Dec 4 '13 at 19:30
[I'm following Turi's lecture notes. Maybe the term was covered in additional exercises or some other supplement.] –  Alice Dec 4 '13 at 19:44