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seen | Jul 27 '11 at 17:50 | |
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Apr
26 |
awarded | Popular Question |
Jul
13 |
comment |
Why Lawvere theories have finite products? and more
Andrej, I quote from nLab article on Lawvere theories: "Remark. For T a Lawvere theory, we are to think of the hom-set T(n,1) as the set of n-ary operations defined by the theory. For instance for T the theory of abelian groups, we have T(2,1)={+,−} and T(0,1)={0}." Why is T(2,1) just {+,-},? Wouldn't it include other 'unnamed' operations as well? |
Jul
13 |
awarded | Scholar |
Jul
13 |
accepted | Why Lawvere theories have finite products? and more |
Jul
13 |
comment |
Why Lawvere theories have finite products? and more
Thanks, after reading Andrej's post, now (3) makes sense. |
Jul
13 |
comment |
Why Lawvere theories have finite products? and more
Thank you, very clear. I was misunderstanding it all, not thinking in syntax at all, just thinking about the models. |
Jul
12 |
awarded | Student |
Jul
12 |
comment |
Why Lawvere theories have finite products? and more
But isn't the monoid described just by the functor that tells what's the set ($F T$), the operation ($F *$) and the neuter ($F e$)? |
Jul
12 |
comment |
Why Lawvere theories have finite products? and more
I think this is the point where I get confused. Why would I need a product of four things? If model for $T$ would be a functor $F$ from $T$ to $Set$, then the product would be $F * : F T^2 \to F T$. For example if $F T = \mathbb{Z}_n$, then $F * : \mathbb{Z}_n^2 \to \mathbb{Z}_n$ would be the product (+) of my monoid (Z_n, +, 0) in Set, wouldn't it? |
Jul
12 |
asked | Why Lawvere theories have finite products? and more |