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I call adjunction algebra a universal algebra with one binary operation denoted as the punctuation sign (;) "semicolon" (but I will be using only one space after it, not on both sides - to avoid going against the regular punctuation practice and creating typographical problems), and 3 axioms which are universal closures of these formulas (between brackets I indicate how I refer to them):

  • $((x; y); y) = (x; y)$ [right idempotency]

  • $((x; y); z) = ((x; z); y)$ [right commutativity]

  • $((x; y); z) = (x; y) \rightarrow ((x; z) = x \vee z = y)$ [right atomicity]

I call normal or normalized adjunction algebra, an algebra which additionally to adjunction has a constant $0$ such that:

  • $(0; x) \ne x$

I think, this algebra is important for algebraization of set theory. Also, for several years now, I am using it, in another presentation, in my research of semantics of natural languages. I used to call the operation (;) "qualification", because it explains the semantics of qualified names in computer science. Lately, I learnt that this operation is called "adjunction" from @Joel's answer to my question. The axioms above are from the paper by Kirby referenced there. True, Kirby does not call this "adjunction algebra", because along these axioms Kirby also uses an induction axiom and he studies a "theory", not an "algebra".

The intended interpretation in set theory of the $(x; y)$ is $x \cup ${$y$}, and in this interpretation $(x; y) = x$ iff $y \ \epsilon \ x$. This shows why adjunction is important for set theory - set theory can be presented in the language of one operation. In Kirby's paper the operation of union is defined by induction (which is an axiom) and it sounds to me improbable that it can be defined through ";" an "0" without such an axiom.

Here are some of my questions:

(1) Does the axioms above occur in abstract algebra and what are the names of the properties they express?

(2) Is what I call normal adjunction algebra "free" in some sense (my algebraic background is not very strong, so my question might not be precise).

(3) Are there algebras where the induction principle, in any of its different forms, is represented algebraically (i.e. as identities or quasiidentities)?

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    $\begingroup$ In right atomicity shouldn't one of the consequents be (x; z) =x ? $\endgroup$ – The Masked Avenger Aug 1 '14 at 21:54
  • $\begingroup$ An example of a non-free normal adjunction algebra: consider the set with 2 elements $\{0, 1\}$, with the operation $;$ defined as $0; 1=0$, $0; 0=1$, $1; 1=1$, $a; b=b; a$. This seems to me to yield a normal adjunction algebra, and it is certainly not free. $\endgroup$ – Noah Schweber Aug 1 '14 at 23:08
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    $\begingroup$ I've deleted a nonsensical comment. Since your axioms for adjunction algebras include a non-equation, there doesn't seem to be a nice way to build a free algebra: "free on $n$ generators" should mean "epis onto anything $n$-generated," but it's not obvious how to build one (or even if there are any), and of course this is even more true for normal adjunction algebras. On the other hand, a finite adjunction algebra certainly won't be free, so I've left my second comment intact. $\endgroup$ – Noah Schweber Aug 1 '14 at 23:20
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    $\begingroup$ How does that change anything? The problem is not in the definition of normality. The problem is that the "right atomicity" axiom is not an equation. $\endgroup$ – Emil Jeřábek Aug 2 '14 at 9:43
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    $\begingroup$ It seems to me like a far more natural notion would be to fix a set $A$ and consider algebras which are sets $P$ together with operations $P\times A\to P$ satisfying (some of) your axioms. If you use only the first two axioms, the free algebra on one generator is the set of finite subsets of $A$. $\endgroup$ – Eric Wofsey Aug 2 '14 at 14:33

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