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Is it true that any homomorphism $f: M \to N$ between two monoids $M$ and $N$ maps generators of $M$ to generators of $N$? I am having trouble proving it to myself.

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    $\begingroup$ The trivial morphism doesn't. $\endgroup$
    – Arnaud D.
    Mar 10, 2020 at 15:06

2 Answers 2

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It has no reason to be true :

If you consider the monoid on one generator, $M = <a>$. Note that $M$ is isomorphic to $\mathbb{N}$, associating to each word $w = aa\ldots a$ the number $n$ of $a$ in that word. We consider morphisms from $M$ to itself, then we can send $a$ to however many $a$'s we want

For instance, if you send $a$ on $a$, you get the identity morphism, it works. But if you chose to send $a$ on $aa$, then you also get a monoid morphism, which associates to each word $w=aa\ldots a$ of length $n$, the word $w'=aa\ldots a$ of length $2n$, containing twice as many $a$'s. Through the previous isomorphism with $\mathbb{N}$. This corresponds to the monoid endomorphism of $\mathbb{N}$ defined by $f(n) = 2n$.

Similarly, if you chose to send $a$ on the word containing $k$ times the letter $a$, then you define the monoid endomorphism of $\mathbb{N}$ $f(n) = kn$.

These are all perfectly valid morphism, and only the identity sends the generators on the generators. And it stays false if you add more generators

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  • $\begingroup$ Do you know what might be meant then by the statement,"every monoid N has an underlying set |N|, and every monoid homomorphism f : N → M has an underlying function |f| : |N| → |M|"? $\endgroup$ Mar 10, 2020 at 21:41
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    $\begingroup$ The underlying set of a monoid is different from its set of generators! In my example, the monoid has only one generator $a$, but its underlying set contains all words you can wrote with only $a$'s. The underlying function is really just a fancy way of saying that a monoid morphism sends all the elements of a monoid onto elements of another monoid $\endgroup$ Mar 10, 2020 at 21:52
  • $\begingroup$ Ahh so the underlying set of a monoid (A,*) is just the set A, ignoring the binary operation? $\endgroup$ Mar 10, 2020 at 22:00
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    $\begingroup$ Yes exactly. Usually we say "forgetting" the binary operation, and it is often referred to as the "forgetful functor", from the category of monoids to the category of sets $\endgroup$ Mar 10, 2020 at 22:02
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Another good reason this isn't true is that a monoid homomorphism is defined with no reference to generators, but monoids can be generated by many different choices of generating sets. For instance, $\mathbb Z$ may be generated either by $\{1\}$ or by $\{-1\}$. Thus every homomorphism $\mathbb Z\to\mathbb Z$ would have to send both $1$ and $-1$ to both $1$ and $-1$!

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  • $\begingroup$ Ahh so generating set is not unique! $\endgroup$ Mar 10, 2020 at 21:37

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