I have as a theorem that the automorphism group of a function is homomorphic to $\Bbb Z/2\Bbb Z$. While it is fairly clear to me what this means im my specific case, I'm unclear what it means in general.
Question:
What are the automorphisms of a function?
The case in particular is as follows:
Let $\Bbb Z_2$ denote the space of $2$-adic integers. Let $f:\Bbb Z_2\to\Bbb Z_2$ be given by $f(x)=(x-1)/2$ if $x$ is odd and $f(x)=x/2$ if $x$ is even. Then the automorphisms of this function are homomorphic to $\Bbb Z/2\Bbb Z$.
This much is obvious to me: The automorphism $\phi:x\mapsto -1-x$ exchanges ones and zeroes, therefore there's a symmetry of the function $f$ which exchanges ones for zeroes in the number's binary representation. Therefore it exchanges the $(x-1)/2$ for the $x/2$ option in the function $f$. This "automorphism" will also be a homeomorphism and an isometry on $\Bbb Z_2$ and therefore it will topologically conjugate $f$ to itself. Moreover, it is a symmetry of the functional graph of $f$. But it's not the only symmetry of that graph. The functional graph of $f$ is the infinite non-rooted binary forest, and it would appear to have many symmetries such as picking any tree within it, and exchanging left for right in that tree alone, or below some point in any tree. None of these, of course, would be homeomorphisms.
I can suggest that the author's intended meaning is that there is only one homeomorphism on $\Bbb Z_2$ which topologically conjugates $f$ to itself. Is that a standard meaning of an "automorphism of a function"? Is it standard in the sense of a function on a topological space? Or is there another definition of automorphism of a function, for which I don't need to invoke the homeomorphism property?
Best Answer
Your assumption is indeed correct. And it is also confirmed, by the way, by the papers that are cited in the paper that you mentioned.
Here is a more general perspective, using category theory.
To every category $\mathcal{C}$ we can associate a new category $\mathrm{Mor}(\mathcal{C})$ of morphisms in $\mathcal{C}$. Objects are morphisms $f : A \to B$, and a morphism from $f : A \to B$ to $g : A' \to B'$ is a commutative diagram.
$$\require{AMScd} \begin{CD} A @>{f}>> B \\ @V{h}VV @VV{h'}V \\ A' @>>{g}> B' \end{CD}$$ An automorphism of $f : A \to B$ is then a pair of automorphisms $\alpha \in \mathrm{Aut}(A)$, $\beta \in \mathrm{Aut}(B)$ such that $\beta \circ f = f \circ \alpha$.
If you only look at endomorphisms, you can also take a different definition: We have a category $\mathrm{End}(\mathcal{C})$ that consists of endomorphisms $f : A \to A$, and a morphism $f : A \to A$ to $g : B \to B$ consists of a single morphism $h : A \to B$ with $h \circ f = g \circ h$.
It follows that (in this category) an automorphism of $f : A \to A$ is just an automorphism $\alpha \in \mathrm{Aut}(A)$ with $$\alpha \circ f = f \circ \alpha,$$ or equivalently $$f = \alpha^{-1} \circ f \circ \alpha.$$
In the paper that you mentioned, $\mathcal{C} = \mathbf{Top}$ is the category of topological spaces and $A$ is the space of $2$-adic integers.