This question refers to an example illustrating properties of the Yoneda lemma: Basic Example of Yoneda Lemma?
Here the paragraph of matter in my question:
By Definition the Yoneda lemma provides a bijection between $Hom(Hom_{C}(-,A), F) $ and $F(A)$, so in this case $Hom(Hom_{M}(-,*), \phi) \cong \phi(*)$.
How do we conclude here that YL provedes the identification of a $M$-action $\phi$ (defined above as functor $M \to Set$) with a function $h: M \to \phi(*)$ with the rule $h(m_1m_2) = \phi(m_1)(h(m_2))$? I don't see where does it come.
It doesn't seem to be true. If there is an $M$-action on $A$ and on $B$, then there is an $M$-action on $A\sqcup B$. Which map $M\to A\sqcup B$ is this $h$?
The Yoneda lemma here says that each $x\in \phi(\star)$ there is a unique map $h_x:M\to \phi(\star)$ so that $h_x(m_1m_2)=\phi(m_1)h_x(m_2)$. That is, $h_x$ is a natural transformation from $\phi$ to $\hom_{M}(-,*)$. This map is $h_x(m)=\phi(m)(x).$
This in turn means that $\phi$ is equivalent to an $h:M\times X\to X$ with the usual definition of an action: $h(m_1,h(m_2,x))=h(m_1m_2,x).$ and $h(1,x)=x$.