Talk:Introduction to Category Theory/Monoids

Isn't the beginning of this a bit scrambled? Addn is first presented as a unary operation, then the equation is written as if it were nullary, I want to define composition as sequential application of the unary version but can't quite make the pieces come together (bad brain day?) AveryAndrews 21:20, 23 November 2007 (UTC)

OK I think I have this the way I want it; of course I may be missing something that your version is supposed to convey 124.168.206.105 00:34, 24 November 2007 (UTC)

Yeah bad brain day i think. Your lambda version defines a function. For arrows $$Addm \circ Addn = Add[m+n]$$ by definition. There is nothing more: no elements to apply to, no results to get from. Tlep 20:38, 25 November 2007 (UTC)

I think I'm still having a bit of a conceptual problem (bad brain month/year/life?), in that I think of the arrows being functions, but where you've just forgotten about the fact that they can be applied to things, i.e. just a set of ordered pairs meeting certain conditions, without the eval operation that defines application. So when it comes time to define composition, you can do it in terms of the binary operation (remembering that there's something there that can be applied to two things), or in terms of the function (remembering that there's something that can be applied to one thing). 124.168.202.162 21:14, 5 December 2007 (UTC)


 * It should not take that long to define a monoid. What is your principal reference?  Hillgentleman|Talk 12:07, 18 February 2008 (UTC)

Proof Clean Up
It seems that the proof (the Monoid of integers is not free) could be improved. First, the first paragraph seems to be ruling out the case where 1 is a generator. It should explicitly assume that M is freely generated by S (and perhaps even mention that this is a proof by contradiction). Then, it is still unclear what the objective of the second part is. It does not seem to make use of the fact that we have already ruled out 1 as a generator. It should also stated that a negative generator need exist, if not give a short explanation. I propose the following in it's place.

Proposition. The Monoid of integers is not free.

Proof.

Assume, by contradiction, that M is freely generated by some set S. Then we proceed by attempting to find two different compositions (linear combinations) of the generators that generates the number 0. Notice that 0 is generated by the composition of zero generators. Moreover, the numbers 1 and -1 must be generated by distinct compositions of generators $$(s_1 + s_2 + ... + s_n)$$ and $$(t_1 + t_2 + ... + t_n)$$. But the composition of these two generators must give zero. But this is a contradiction, since S is freely generated by S. Therefore, there does not exist a subset of M such that S freely generates M, and thus M is not free. Brent Perreault (talk) 04:12, 17 December 2012 (UTC)