tryingotounderstand

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seen Jul 27 '11 at 17:50

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