# Taking the (pseudo)inverse of a monoid operation.

Let $M$ be a monoid with binary operation $f : M \times M \to M$. I'm interested in functions $g : M \to M\times M$ that obey the property:

$$f(g(m)) = m$$

I want to understand what all of the possible $g$ functions can look like under different monoids. So my question is where can I find more information about these structures? Also, what should I call a monoid with a designated pseudoinverse for the binary operation?

Some background that I've worked out so far:

It is easy to see that the function $g$ is not unique in any monoid with more than one element. For example, the functions

\begin{align*} g_1(m) &= (\epsilon,m) \\ g_2(m) &= (m,\epsilon) \end{align*}

are pseudoinverses for every monoid.

It seems like these functions might have a number of interesting forms depending on the structure of the monoid. Here's three different examples:

1. The rationals under addition allow a function $g(x) = (x/2,x/2)$. No matter how many times we successively apply $g$, we will never reach the identity.

2. The naturals under addition do not have a similar inverse that can be applied indefinitely. Because it is cyclic, we will eventually decompose into ones and zeros.

3. The naturals under multiplication is not cyclic, so we will eventually decompose into the prime factors of a number.

• What means "we successively apply $g$"? – Boris Novikov Dec 16 '13 at 22:48
• $g(x)=(x/2,x/2); g(x/2)=(x/4,x/4); g(x/4) = (x/8,x/8)$ and so on. I just mean that we can apply $g$ to its left output, right output, or both outputs. – Mike Izbicki Dec 16 '13 at 23:00
• The $g$ functions calculates two divisors of a monoid element, and by iterating you are calculating the irreducibles (if any) dividing that monoid element. These topics are all fairly well-understood. – vadim123 Dec 16 '13 at 23:09
• You would love Hopf monoids. – Berci Dec 16 '13 at 23:21
• @Berci I don't think this can be a Hopf monoid. Using rationals under addition as the operation, I'll take the pseuodinverse $g(x)=(x/2,x/2)$ as above. But $(g(x),x) \ne (x,g(x))$ so the coassociativity law doesn't hold. Of course, I might be misunderstanding something. – Mike Izbicki Dec 17 '13 at 0:00