A morphism \\(f:F(m)\rightarrow n\\) is an element \\(f \in N\\) such that \\(f\circ_N F(m) = n\\).

Likewise, a morphism \\(g:m\rightarrow G(n)\\) is an element \\(g \in M\\) such that \\(g\circ_M m = G(n)\\).

Thus we have \\(g \circ_M m = G(n) = G(f \circ_N F(m)) = G(f) \circ_M G(F(m))\\). If we consider the case \\(m = 1_M\\) and recall that \\(G(F(1_M))=G(1_N)=1_M\\), this becomes \\(g=G(f)\\) so we have the mapping in one direction.

In the other direction, using \\(g=G(f)\\) and \\(n = 1_N\\), I get the condition \\(g \circ_M G(F(m)) = g \circ_M m = 1_M\\), i.e., for any \\(m \in M\\) we must have \\(g \circ_M m = 1_M\\) if and only if \\(g \circ_M G(F(m)) \\). So for all elements \\(m \in M\\) that have a left-inverse \\(g\\), this must also be the left-inverse of \\(G(F(m))\\). This would surely work if \\(G\\) was a left-inverse of \\(F\\), but it is not clear to me that it is necessary - nor what monoid concept this corresponds to.

Likewise, a morphism \\(g:m\rightarrow G(n)\\) is an element \\(g \in M\\) such that \\(g\circ_M m = G(n)\\).

Thus we have \\(g \circ_M m = G(n) = G(f \circ_N F(m)) = G(f) \circ_M G(F(m))\\). If we consider the case \\(m = 1_M\\) and recall that \\(G(F(1_M))=G(1_N)=1_M\\), this becomes \\(g=G(f)\\) so we have the mapping in one direction.

In the other direction, using \\(g=G(f)\\) and \\(n = 1_N\\), I get the condition \\(g \circ_M G(F(m)) = g \circ_M m = 1_M\\), i.e., for any \\(m \in M\\) we must have \\(g \circ_M m = 1_M\\) if and only if \\(g \circ_M G(F(m)) \\). So for all elements \\(m \in M\\) that have a left-inverse \\(g\\), this must also be the left-inverse of \\(G(F(m))\\). This would surely work if \\(G\\) was a left-inverse of \\(F\\), but it is not clear to me that it is necessary - nor what monoid concept this corresponds to.