Free monoids as universal constructions (example)

I'm a beginner and I'm desperately trying to understand free monoids as universal constructions using an example.

What is the relation between the following?

• a generator set (e.g. $$\mathbb{N}$$ (natural number set))

• the free monoid $$m$$ where the only "equations" / identifications must be those that follow from the monoid laws (e.g. using concatenation)

• a non-free monoid $$n$$ where of course we have extra "equations" / identifications (like $$(N, +, 0)$$ with $$1 + 2 = 3$$)

Is the unique morphism from $$m$$ to $$n$$ meant to show that those extra equations exist?

What about forgetful functors from these two monoids? Are they from the monoid to the generator set or to the actual set of generated elements?

If we have extra equations in the non-free monoid, how cam there be a morphism between the co-domains of the above forgetful functors?

Some clarifications My intention in the above mess is to get an example of an example of a free monoid using universal construction (if such definitions are possible).

My understating was that in order to do that you need multiple related monoids (as in any universal constructions), and the following "ingredients": functors from monoids to their underlying sets, functions form the generator set to those underlying sets and functions between the underlying sets. The details are not clear to me, so I regret the messy "question" :(

• $1+2=3$ isn't an "extra equation". $\mathbb{N}$ is the free monoid on one generator. – Malice Vidrine Oct 19 '18 at 16:54
• I hope I didn't come across as too critical; I didn't mean to suggest it was a bad question, so much as to say "due to the nature of being confused about something, I worry that my answer may not say the things you would find most helpful." If I may ask, where are you learning about free monoids and universal constructions from? Perhaps if I know the source material I can see where things are getting muddled. – Malice Vidrine Oct 21 '18 at 18:30

1) The generating set is just a set. There's not too much to say about it except that the point of the Free/Forgetful adjunction (using $$FS$$ for the free monoid on $$S$$ and $$UM$$ for the underlying set of a monoid $$M$$) is that there are exactly as many monoid homomorphisms $$FS\to M$$ as there are ordinary functions $$S\to UM$$ (and moreover this bijection occurs in a nice way).

Note that the generating set $$S$$ is not necessarily the same as $$UFS$$, so it's important that these are clearly distinguished ($$\mathbb{N}$$ is the underlying set for $$(\mathbb{N},0,+)$$, but $$1$$ is its generator).

2 & 3) The previous point tells you most of what there is to say about fee monoids, but let's compare, for instance, the free monoid on one generator, $$(\mathbb{N},0,+)$$, to another monoid generated by a single element: the cyclic group $$\mathbb{Z}_2$$ (all groups are also monoids).

It's obvious that in both cases, any element can be written as a sum of the right number of $$1$$s. But for $$\mathbb{Z}_2$$ you can only sum $$1$$ so many times before you start hitting a number you've already hit. In contrast, in the free monoid, you can sum $$1$$ as many times as you like and you will always get a new element. That's because, if you construct the free monoid as concatenation of strings, there is nothing about concatenation that says "concatenating too many things results in an empty string".

After your bulleted points, I want to address a few misconceptions that seem to be expressed.

First, there are typically many morphisms from a free monoid to an arbitrary monoid---even from $$\mathit{FUM}$$ to $$M$$---so talking about "the unique" morphism $$m\to n$$ above doesn't make sense. There is a special morphism $$\varepsilon:\mathit{FUM}\to M$$, with the property that for any $$f:FS\to M$$ there's a unique $$j:S\to UM$$ with $$\varepsilon\circ Fj=f$$.

The forgetful functor takes a monoid to its underlying set, not its generators. But the monoids do not "have forgetful functors" in any way that makes sense for the free/forgetful adjunction; there is a single forgetful functor from the category of monoids to the category of sets. As such your last question doesn't actually make sense. There can be a morphism $$UM\to UM'$$ for any two monoids $$M,M'$$ because such morphisms are just functions, and $$UM,UM'$$ are always non-empty, so there's always at least one function. I suspect you mean something different, but I can't tell what that is.

• The question has, really, a lot of questions to clear up, and some of them are predicated on misunderstandings, so I've done the best I can. Let me know if I've misunderstood something. – Malice Vidrine Oct 20 '18 at 1:48
• Thank you very much for trying to make sense of my non sense... – jack malkovick Oct 21 '18 at 15:11
• @jackmalkovick - No problem, there are several big ideas that come into play in free constructions, so it can be easy to get lost. I hope my effort to untangle some of them is at least a little helpful. – Malice Vidrine Oct 21 '18 at 18:16