Generalized free product of semigroups with amalgamated subsemigroups Hanna Neumann in 
[American Journal of Mathematics, 1948, 
http://www.jstor.org/discover/10.2307/2372201?uid=2&uid=4&sid=21102497379451 ] 
introduced a notion of generalized free product of groups $G_i$ with amalgamated subgroups $H_{ij}< G_i$.  Did anybody consider this construction for semigroups? Thank you in advance. 
This question arose while studying partial actions of groups on semigroups.
 A: OK, here is my comment expanded to an answer. I looked at Hanna Neumann's definition again. What she defined is not a special case graph of groups, but of what is now called (again, in a special case) the "fundamental group of a complex of groups". Indeed, the modern definition of the latter is category-theoretic one (the limit of a commutative diagram of groups), details could be found for instance in the book by Bridson and Haefliger "Metric Spaces of Non-Positive Curvature". This definition extends verbatim to the case of semigroups. It is then well-known that vertex groups in such complex do not (in general) embed in the limit. However, they do provided that the complex is developable, which, in turn, is implied for instance by the local CAT(0) assumption. 
Search for "complex of semigroups" returned nothing. My guess then is that you are on your own. My suggestion is to look for a CAT(0)-concept which might work in this context and would yield a "developable complex" of semigroups and guarantee, say, that vertex semigroups are embedded. Reading first Stallings' paper "Triangles of groups" (where "angles" are defined purely algebraically) might be a good start. See also here and Bridson-Haefliger.  
A: If you allow many amalgamated subsemigroups, then the notion is too general. Every finitely presented semigroup can be presented that way with all $G_i$ free semigroups. For example, take any semigroup with one defining relation $S=\langle X\mid u=v\rangle$. Then consider a copy $X'$ of $X$, with a bijection $x\mapsto x'$, and two free semigroups $F(X), F(X')$. Identify subsemigroup $\langle x\rangle$ of $F(X)$ with $\langle x'\rangle$ of $F(X')$ for every $x\in X$, and $\langle u\rangle$ with $\langle v'\rangle$. Then you get $S$. Similarly, you can get any finitely presented semigroup, and if you allow infinite number of amalgamated subsemigroups - any semigroup. 
A: The main difference between amalgamating groups and semigroups is that 
amalgamation is not always possible for semigroups.
One of the simplest examples (from Lyapin's book) is the following.
Take two commutative semigroups (which are even groups):
$$
A=\{\dots,{1\over16},{1\over8},{1\over4},{1\over2},1,2,4,\dots\}
\quad\hbox{and}\quad
B=\{\dots,{1\over16},-{1\over8},{1\over4},-{1\over2},1,2,4,\dots\},
$$
where the operation in $A$ is the usual multiplication and the operation in
$B$ is $x\circ y=\pm xy$, where the sign is chosen so that $x\circ y\in B$.
It is easy to see that the amalgam (where 
$A\cap B=\{\dots,{1\over16},{1\over4},1,2,4,\dots\}$) 
cannot be embedded in a common semigroup. Indeed, in such a common semigroup, we would have
$$
{1\over2}={1\over2}\cdot1={1\over2}\cdot(2\circ(-{1\over2}))=
({1\over2}\cdot2)\circ(-{1\over2})=1\circ(-{1\over2})=-{1\over2}.
$$  
