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The forgetful functor $U : \mathsf{Ab} \to \mathsf{Set}$ is monadic, this follows from Beck's monadicity theorem or some other general result. Anyway, I would like to prove this directly, thereby solving an exercise in Mac Lane's Categories for the working mathematician. This amounts to prove an equivalent definition of an abelian group, using $\mathbb{Z}$-linear combinations of arbitrary length, which should be quite elementary. But working through the details, I get stuck at some point.

First, let me describe the monad $T: \mathsf{Set} \to \mathsf{Set}$ corresponding to $U$. It maps a set $X$ to the set $T(X)$ of functions $\lambda : X \to \mathbb{Z}$ with finite support. If $f : X \to Y$ is a map, then $T(X) \to T(Y)$ sends $\lambda$ to the function $y \mapsto \sum_{x \in f^{-1}(y)} \lambda(x)$. The unit $\eta : \mathrm{id} \to T$ sends $x \in X$ to the function supported at $x$ with value $1$. The multiplication $\mu : T^2 \to T$ maps $\alpha \in T^2(X)$ to $\mu(\alpha) \in T(X)$ defined by $\mu(\alpha)(x)=\sum_{\lambda \in T(X)} \alpha(\lambda) \cdot \lambda(x)$.

It follows that a $T$-module may be described as follows: This is a set $X$ together with a map $h : T(X) \to X$, which I will denote by $f \mapsto \sum_{x \in X} f(x) \cdot x$, such that the following properties hold:

1) If $f \in T(X)$ is supported at $x_0 \in X$ with value $1$, then $\sum_{x \in X} f(x) \cdot x = 1$.

2) For every $\alpha \in T(T(X))$ we have an equality

$$\sum_{x \in X} \Bigl(\sum_{\lambda \in T(X)} \alpha(\lambda) \cdot \lambda(x)\Bigr) \cdot x = \sum_{x \in X} \Bigl(\sum_{\lambda \in T(X), \sum\limits_{y \in X} \lambda(y)=x} \alpha(\lambda)\Bigr) \cdot x$$

From the notation it is clear how the comparison functor $\mathsf{Ab} \to \mathsf{Mod}(T)$ looks like. Conversely, let $(X,h)$ be a $T$-module. Then I would like to define a binary operation $+$ on $X$ roughly by $x+y = 1 \cdot x + 1 \cdot y + 0 \cdot ? + \dotsc$. Unfortunately, the precise definition requires a case distinction! First, define $0_X := \sum_{x \in X} 0 \cdot x$. Then, for $x,y \in X$, define

$$x+y :=\begin{cases} \sum_{z \in X} \delta_{z,\{x,y\}} \cdot z & x \neq y\\ \sum_{z \in X} (2 \cdot \delta_{z,x}) \cdot z & x = y \neq 0_M \\ 0_M & x=y=0_M \end{cases}$$

With such a definition, it seems to be very hard to prove associativity. Or is there an elegant proof? Or a better definition of $+$?

Martin Brandenburg
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1 Answers1

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The general strategy for these things is to regard $T (X)$ as being the set of all terms in the language of the algebraic theory in question, with variables drawn from $X$, modulo provable equality in that algebraic theory. (I like to think this is the reason why monads are denoted by $T$; but probably the real reason is in the now-outdated name ‘triple’.) From this point of view, the action map $h : T (X) \to X$ interprets these formal terms as actual elements in $X$ in the obvious way.

In particular, $T (X)$ will always have the structure of a $T$-algebra, and $h : T (X) \to X$ will always be a surjective $T$-algebra homomorphism, with a (set-theoretical) splitting $\eta_X : X \to T (X)$. Thus, the least painful way of defining the algebraic structure on $X$ is by lifting the elements of $X$ to $T (X)$ via $\eta$, using the algebraic operations of $T (X)$, and descending back to $X$ via $h$.

For example, if we define $T (X)$ to be the free abelian group on $X$ in the way you do, $T (X)$ is very easily seen to be an abelian group, so we can define the addition on $X$ by $x + y = h(\eta_X(x) + \eta_X(y))$. It now follows that $+$ is a unital commutative associative binary operation with two-sided inverses. For example, \begin{align} x + (y + z) & = h(\eta_X(x) + \eta_X(h(\eta_X(y) + \eta_X(z)))) \\ & = h(\eta_X(h(\eta_X(x))) + \eta_X(h(\eta_X(y) + \eta_X(z)))) \\ & = h(T h (\eta_{T X}(\eta_X(x))) + T h (\eta_{T X}(\eta_X(y) + \eta_X(z)))) \\ & = h(T h (\eta_{T X}(\eta_X(x)) + (\eta_{T X}(\eta_X(y) + \eta_X(z))))) \\ & = h(\mu_X (\eta_{T X}(\eta_X(x)) + (\eta_{T X}(\eta_X(y) + \eta_X(z))))) \\ & = h(\mu_X (\eta_{T X}(\eta_X(x))) + \mu_X (\eta_{T X}(\eta_X(y) + \eta_X(z)))) \\ & = h(\eta_X (x) + (\eta_X (y) + \eta_X (z))) \\ & = h((\eta_X (x) + \eta_X (y)) + \eta_X (z)) \\ & \qquad\vdots \\ & = (x + y) + z \end{align}

Zhen Lin
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  • Thank you! Is it really immediate that $+$ is a group law? – Martin Brandenburg Jan 13 '13 at 00:21
  • Sure – we can check the equations in $T (X)$ using $\eta$ and $h$. – Zhen Lin Jan 13 '13 at 08:12
  • For example, $x+(y+z) = h(\eta(x)+\eta(h(\eta(y)+\eta(z))))$, why does this simplify to $h(\eta(x)+\eta(y)+\eta(z))$? – Martin Brandenburg Jan 13 '13 at 10:43
  • I added a postscript. We need to use the $T$-algebra axioms and lift to $T(T X)$ at one point. – Zhen Lin Jan 13 '13 at 12:16
  • Thanks you. In one step, you use that $\mu_X : T(T(X)) \to T(X)$ is a homomorphism of abelian groups; this follows easily from the description I gave above. But you also seem to use that $T(h) : T(T(X)) \to T(X)$ is a homomorphism of abelian groups. Don't we need to prove this? – Martin Brandenburg Jan 13 '13 at 15:22
  • Yes, I did assume that. But $T f$ for any function $f$ is quite easily seen to be a homomorphism of abelian groups, especially since it sends generators to generators. – Zhen Lin Jan 13 '13 at 15:50
  • Ok you are right, this also follows from the description I gave above. Sorry for these stupid questions. After all, your proof has the same ideas as the proof of the general Beck's monadicity Theorem, right? – Martin Brandenburg Jan 13 '13 at 17:21
  • Hmmm, yes and no. The argument here is the usual one used to show that the category of models for an algebraic theory satisfies the split coequaliser condition of Beck's monadicity theorem, but it is not crucial in the theorem itself. – Zhen Lin Jan 13 '13 at 18:46