How can I show that every monoid $M$ admits a surjection from a free monoid $F(X) \rightarrow M$ ?

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    $\begingroup$ Have you tried taking the free monoid on M itself? $\endgroup$ – Robert Furber Nov 25 '16 at 20:55
  • $\begingroup$ Universal properties are your friend (for a lifetime). $\endgroup$ – HeinrichD Nov 25 '16 at 23:58

The set $X$ is called an alphabet. You can take $M$ itself as alphabet and then the morphism $\mu : F(M)\rightarrow M$ is just the multiplication.

To be more specific, the structure of monoid is the data of a triplet $(M,*,1_M)$ where

  • $M$ is a set
  • $*$ is an associative internal law in $M$,
  • $1_M$ is the neutral

The elements of $F(M)$ are strings $m_1.m_2.\cdots .m_k$ where the $m_i\in M$, then $$ \mu(m_1.m_2.\cdots .m_k)=m_1*m_2*\cdots *m_k $$ the stars standing for the multiplication inside the monoid $M$.

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    $\begingroup$ Of course this is the right answer, but should we be answering such non-research-level questions on MO? $\endgroup$ – LSpice Nov 25 '16 at 21:04
  • $\begingroup$ Also you can take for X a generating set for M. Since the algebra has only one binary operation (and one constant operation), the generating set can be significantly smaller than the underlying set for M only if the underlying set is countable. Gerhard "This Is Basic General Algebra" Paseman, 2016.11.25. $\endgroup$ – Gerhard Paseman Nov 25 '16 at 21:07
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    $\begingroup$ Someone should answer the questions. Someone else should migrate the question. Those who migrate should edit the post to indicate why the migration occurred. Gerhard "We Can Work Outside MathOverflow" Paseman, 2016.11.25. $\endgroup$ – Gerhard Paseman Nov 25 '16 at 21:12
  • $\begingroup$ what do you mean by "the stars standing for the multiplication inside the monoid M" ? $\endgroup$ – Daniel Nov 25 '16 at 23:09
  • $\begingroup$ Just, I note $(M,*,1_M)$ the monoid. See the amended answer. $\endgroup$ – Duchamp Gérard H. E. Nov 26 '16 at 1:40

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