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Right now I'm working on a topological graph theory problem. To prove a theorem I introduced some objects. Has anyone heard of something similar before? I would like to call them by the right name.

Fix a space $X$. Our objects are pairs $(f,U)$ where $f\colon X\to X$ and $U\subset X$. Multiplication is given by $$ (f,U)\rtimes (g,V)=(g\circ f,U\cap f^{-1}V). $$ These objects form a monoid with identity $(\text{id},X)$.

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  • $\begingroup$ Aren't these just partial functions? $\endgroup$ Commented Oct 24, 2014 at 4:07
  • $\begingroup$ @NoahS What is "partial" about $f:X\to X$? $\endgroup$
    – bof
    Commented Oct 24, 2014 at 4:21
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    $\begingroup$ If $f$ was defined only on $U$ it would be a function "partially" defined on $X$. I suspect that's what he means. Your monoid structure is the natural composition of function operations in that setting. $\endgroup$ Commented Oct 24, 2014 at 4:46
  • $\begingroup$ Actually, it is crucial in my application each $f$ is a homeomorphism of a topological space, and therefore not a partial function at all. Even without this specific context, the pairs $(f,U)$ and $(f,V)$ are treated as distinct objects - even when the function $f$ has the same domain and codomain in both pairs. $\endgroup$
    – pre-kidney
    Commented Oct 25, 2014 at 2:56

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I believe this is an instance of a semidirect product of one monoid acting on other, construction intermediate in generality between semidirect product of groups and Grothendieck construction/fibration of a 2-functor.

For completeness, I will reproduce the definitions: semidirect product $M\ltimes N$ of a group $M$ and a group $N$ acted upon (say, from the right) by $M$ is their product with the group operation $(m,n)(m',n')=(mm',n^{m'}\cdot n')$. It can also be characterized as the universal group with homomorphhic images of both $M$ and $N$ where the action becomes conjugation, i. e. $n\cdot m'=m'\cdot n^{m'}$ for all $n\in N$, $m'\in M$.

The above I formulated in such way that it makes sense for monoids too. This generalization must be well known anyway, maybe directly for monoids but what I know for sure is that it is a particular case of a split fibration. Given a category $M$ and a (say, contravariant) functor $N:M^{\mathrm{op}}\to\textrm{Categories}$, the associated (split) fibration is a category $M\ltimes N$ together with a functor to $M$ given as follows: objects of $M\ltimes N$ are pairs $(X,x)$ where $X$ is an object of $M$ and $x$ is an object of $N(X)$. A morphism from $(X,x)$ to $(X',x')$ is a pair $(f,U)$ where $f:X\to X'$ and $U:x\to N(f)(x')$. And for further $g:X'\to X''$, $V:x'\to N(g)(x'')$ the composite $(g,V)\circ(f,U)$ is $(g\circ f,N(f)(V)\circ U)$.

The latter construction clearly subsumes the former one when $M$ has a single object $X$ and $N(X)$ also has single object.

In your example $M$ is the full subcategory of sets on single object $X$ and $N$ assigns to a set $X$ its powerset considered as a single object category with composition given by intersection, while on morphisms $N$ is given by inverse image.

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