Is this definition equivalent to the definition of clone of functions?

Question

A clone of functions on a set $S$ is a collection of n-ary functions on $S$ which contain projection functions and is closed under composition of functions. But I have a different definition. My definition of a clone is a collection of functions which contain the identity function from $S$ to $S$, is closed under composition of functions, and is closed under adding dummy variables. Are these two definitions equivalent?

Answer 1

Yes, these are equivalent.

To show that every clone satisfies your properties, we just have to show that a clone is closed under adding dummy variables. We can do this via the projection operations; for example, given a binary function $f(u,v)$ in a clone $\mathcal{C}$, we can form the "dummy extension" $$g:(x,y,z)\mapsto f(x,z)$$ as $g=f\circ(\pi_1^3,\pi_3^3)$ where $\pi^i_j$ is the $i$-ary function projecting onto the $j$th coordinate. Its' a quick exercise to prove the fully general result.

In the other direction, we just need to show that all the projection functions can be gotten by "dummification." But this is basically trivial: the definition of a projection function is basically "a "dummification" of the identity."

---

There is a minor subtlety here: we need to allow arbitrary variable insertion in "dummification" constructions. E.g given a binary function $f(u,v)$ we need to allow each of the following as a "dummification" of $f$: $$(x,y,z)\mapsto f(x,y),\quad (x,y,z)\mapsto f(x,z),\quad (x,y,z)\mapsto f(y,z).$$ (Actually we can get by with just "dummification on the left" and "dummification on the right," but still.) One alternate approach would be to just demand closure under one form of "dummification," and separately demand closure under permutation of variables.