Concluding about the relation of two monoids to a third a relation between the two monoids

Question

Let $N, S, T$ be monoids and let $\varphi : N \to S$ and $\varphi' : N \to T$ be surjective homomorphisms such that for each $P \subseteq S$ there exists some $Q \subseteq T$ such that $$ \varphi^{-1}(P) = \varphi'^{-1}(Q). $$ Then there exists a surjective mapping from a submonoid of $T$ onto $S$.

Answer 1

Let's temporarily forget about monoids and start with a claim that is purely about sets:

Claim 1 - Let $\varphi\colon N\to S$ and $\psi\colon N\to T$ be surjective functions such that for each $P\subseteq S$ there is some $Q\subseteq T$ satisfying $\varphi^{-1}(P)=\psi^{-1}(Q)$. Then there is a surjective function from $T$ to $S$.

To prove Claim 1 we need to come up with a function $T\to S$. So given $t\in T$, where shall we send it to in $S$? Since $\psi$ is surjective we know that $t=\psi(n)$ for some $n\in N$, so perhaps we could send $t$ to $\varphi(n)$.

However, we are only allowed to do this (i.e., this will only define a function $T\to S$) if $\varphi(n)$ does not depend on the choice of $n$. We therefore must check that if $t=\psi(m)$ for some $m\in N$ as well then actually $\varphi(m)=\varphi(n)$. This is really the essence of the question, and I don't want to completely spoil it for you, so I won't give a proof of this bit yet.

Now let's write $\chi\colon T\to S$ for the function we have just defined. The key property to remember about $\chi$ is that $\chi\bigl(\psi(n)\bigr)=\varphi(n)$ for all $n\in N$.

To finish the proof of Claim 1 we need to check that $\chi$ is surjective. Since $\varphi$ is surjective, each $s\in S$ can be written as $s=\varphi(n)$ for some $n\in N$, and therefore $\chi$ sends $\psi(n)\in T$ to $s$. So yes, $\chi$ is surjective. Hence we have proved Claim 1. We can now deal with monoids:

Claim 2 - If $\varphi$ and $\psi$ are homomorphisms between monoids then $\chi$ is also a homomorphism.

This follows quickly from our definition of $\chi$. For example, to show that $\chi(1)=1$ we simply observe that we have $$\chi(1)=\chi\bigl(\psi(1)\bigr)=\varphi(1)=1,$$ because $\varphi$ and $\psi$ are homomorphisms. The other property of $\chi$ being a homomorphism - namely that $\chi(t_1t_2)=\chi(t_1)\chi(t_2)$ for all $t_1,t_2\in T$ - follows in a similar way.