4
$\begingroup$

Let $\bf{N}$ be the category whose objects are all the nonegative integers while the morphisms $m\longrightarrow n$ are the mappings from $m$ to $n$, considered as finite sets (so $\bf{N}$ is the skeleton of the category of finite sets and mappings). In $\bf{N}$ we have an operation of coproduct or sum, $m+n$, represented by the disjoint union of sets. In particular, each $n\in Obj\bf{N}$ can be obtained as a sum of $n$ terms: $n=1+1+\ldots +1$.

By an algebraic theory one means a category $\mathscr{P}$ containing $\bf{N}$ as a subcategory, with the same objects and coproducts maps as $\bf{N}$. Thus $\mathscr{P}$ differs from $\bf{N}$ in having (possibly) more morphisms.

Now a $\mathscr{P}$-algebra is a controavariant functor $F$ from $\mathscr{P}$ to $Set$, which converts coproducts to products. Thus $Fn=A^n$, where $A^n$ is the product of $n$ copies of $A^1=A$, and the $\mathscr{P}$-morphisms $1\longrightarrow n$ define mappings $A^n\longrightarrow A$, i.e. $n$-ary operations on $A$.

Given an algebraic theory $\mathscr{P}$, we can form the category $Funct(\mathscr{P}^{op},Set)$ of all product preserving functors from $\mathscr{P}^{op}$ to $Set$.

QUESTION 1: how can I define a forgetful functor $U:Funct(\mathscr{P}^{op},Set)\longrightarrow Set$, associating to each $\mathscr{P}$-algebra its "underlying" set, in such a way that this functor $U$ has a left adjoint $F$ providing the free $\mathscr{P}$-algebra on a given set?

QUESTION 2: can you suggest me some references (books, articles, ...) where I can find a complete description of this topic?

$\endgroup$
  • $\begingroup$ @berci thank u, where can I find the most commonly used definition of algebraic theory? $\endgroup$ – Danae Kissinger Jun 8 '15 at 10:18
4
$\begingroup$

Q1: You already sketched the structure. A functor $F:\mathscr P^{op}\to\mathcal{Set}$ plays the role of an algebra on set $A=F1$ with projections and possible operations $A^n\to A$.
This suggests that the forgetful functor is going to be $F\mapsto F1$.

Q2: It seems, this is just the dual approach to the same algebraic theories, compared to Lawvere theories.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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