Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.


Have I even understood the question properly? Please help. I'd prefer HINTS ONLY.

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.

share|improve this question
    
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 :/ –  Shaun Nov 29 '13 at 17:34
1  
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
1  
Yes, you can do that. –  Zhen Lin Nov 29 '13 at 17:57
1  
The product in $\mathbb{N}^\mathrm{op}$ is the coproduct in $\mathbb{N}$, which is addition. –  Zhen Lin Dec 1 '13 at 22:41

1 Answer 1

up vote 3 down vote accepted
+50

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.

share|improve this answer
    
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! –  Shaun Dec 4 '13 at 11:58
1  
@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 $A^{n}\times\bullet\cong A^{n}$.] –  Shaun Dec 4 '13 at 19:23
    
Oh, I see; it's a bit of a misnomer. –  Shaun 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.] –  Shaun Dec 4 '13 at 19:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.