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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1$\begingroup$ The trivial morphism doesn't. $\endgroup$– Arnaud D.Mar 10, 2020 at 15:06
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2 Answers
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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
answered Mar 10, 2020 at 16:28
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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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1$\begingroup$ 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
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$\begingroup$ 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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1$\begingroup$ 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 $-1$!
answered Mar 10, 2020 at 18:19