Yoneda Lemma for Monoids I have a question about the red tagged statement in the image below. 

Why the morphisms of right $M$-sets are in bijection with elements of $S$? According to Yoneda we have the bijection between natural trafos $Hom(Hom_{BM^{op}}(-,x), F)$ and $F(x)$. Futhermore there is mentioned above that that every presheaf (here $F$ and $Hom_{BM^{op}}(-,x)$) corresponding to $M$-sets. 
So I conclude that $Hom(Hom_{BM^{op}}(-,x), F) \cong Hom_{M-set}(M, S)$ where $S$ is a $M$-set corresponding to presheaf $F$.
My question is why $S = F(x)$ holds?
Whole article: https://qchu.wordpress.com/2012/04/02/the-yoneda-lemma-i/
 A: Presumably you understand that the content of the claim that presheaves on $BM$ are "precisely right $M$-actions" is that these are equivalent (isomorphic, even) categories. Let's call the category of presheaves $\widehat{BM}$ and the category of $M$-actions $\mathbf{Mact}$. This means there are functors $F:\widehat{BM}\to\mathbf{Mact}$ and $G:\mathbf{Mact}\to\widehat{BM}$ with $GF\simeq id_{\widehat{BM}}$ and $FG\simeq id_{\mathbf{Mact}}$. It should then be clear, if you understand the construction of $G$ that $GM\simeq \hom(-,*)$, and that $\widehat{BM}(\hom(-,*),GS)\cong\mathbf{Mact}(M,S)$ follows immediately from the equivalence of categories.
The only missing ingredient to make the claim work is that $GS(*)\simeq |S|$ (where I write $|S|$ for the underlying set of the $M$-action S). This doesn't follow (in any way immediately obvious to me) from some generic property like Yoneda or equivalence with presheaves; this is purely a consequence of how $M$-actions have underlying sets, and the construction of $G$. Where $\phi:|S|\times M\to|S|$ is the map that makes the $M$-action on $S$, we define $GS$ as the presheaf on $BM$ such that $GS(*)=|S|$ and $GS(m)=s\mapsto\phi(s,m)$. Moreover, given $A\in\widehat{BM}$, we can define $FA$ such that $|FA|=A(*)$ and $\phi_A:A(*)\times M\to A(*)$ is $(x,m)\mapsto A(m)(x)$. I leave it to you to check that these are functorial. From this it should be obvious why an equivalence of the two categories means that $\mathbf{Mact}$ morphisms $M\to S$ correspond to elements of $|S|$ in a Yoneda-like way.
