1
$\begingroup$

A monoid $M$ satisfies the universal mapping property (UMP) over $X$, if $X \subseteq M$ and for every map $\varphi : X \to N$, where $N$ is another monoid, there exists a unique homomorphism $\varphi : M \to N$ (i.e. this could be thought of as the extension of $\varphi : X \to N$, but this I want to show).

Now I want to show that if $M$ satisfies UMP over $X$, then $M$ is generated by $X$. First I thought that if this is not the case, then there exists some $m \in M$ such that $m$ is not a finite product of some elements from $X$, and this would contradict uniqueness of the homomorphism, as we could map $m$ onto any element from $X$, or to $1$. But in the first place this just gives a map, and it is not clear that this is also a homomorphism (see this related question).

So how to show that $X$ generates $M$?

$\endgroup$
4
$\begingroup$

First choose $N=\langle X\rangle$, the submonoid of $M$ generated by $X$. We obtain (a unique) $\varphi_0:M\to\langle X\rangle$

Now choose $N=M$ and consider $\varphi_1:M\overset{\varphi_0}\to\langle X\rangle\hookrightarrow M$ and $\varphi_2:=id:M\to M$. Both are homomorphisms and extend the inclusion $X\hookrightarrow M$, hence $\varphi_1=\varphi_2$ by uniqueness, which proves the claim since $im\varphi_1=\langle X\rangle$.


By the way, this property also implies that $M$ must be isomorphic to the free monoid $X^*$ over $X$ which consists of finite sequences of $X$ with concatenation as operation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.