Does this theorem/collection of theorems have a name? Or is it just seen as obvious?

Question

After some thought I realized that given any sets $X$ and $Y$ with equal cardinality and an arbitrary bijection $f:X\to Y$ we get that the set of all bijections from $X$ to $Y$ is given by:

$$\{f\circ \sigma:\sigma\in \text{Sym}(X)\}=\{\sigma\circ f:\sigma\in \text{Sym}(Y)\}$$

Then when I recently learned about the definition of isomorphism/automorphism between sets equipped with binary operations i.e. $(X,*)\cong (Y,\cdot)\iff f(x*y)=f(x)\cdot f(y)$ for some bijection $f:X\to Y$, I found that the set of all isomorphisms from $(X,*)$ to $(Y,\cdot)$ is given by:

$$\{f\circ \sigma:\sigma\in \text{Aut}(X)\}=\{\sigma\circ f:\sigma\in \text{Aut}(Y)\}$$

Where $\text{Aut}(X)$ and $\text{Aut}(Y)$ are the set of all automorphisms on $(X,*)$ and $(Y,\cdot)$ respectively. Well it turns out this holds for even more general algebraic structures with operations of arbitrary arity using this definition https://en.wikipedia.org/wiki/Homomorphism#Definition i.e. if $(X,\mu_X)\cong (Y,\mu_Y)\iff f(\mu_X(x_1,\ldots x_n))=\mu_Y(f(x_1),\ldots f(x_n))$ for a bijection $f:X\to Y$ then we again get:

$$\{f\circ \sigma:\sigma\in \text{Aut}(X)\}=\{\sigma\circ f:\sigma\in \text{Aut}(Y)\}$$

Where $\text{Aut}(X)$ and $\text{Aut}(Y)$ are the set of all automorphisms on $(X,\mu_X)$ and $(Y,\mu_Y)$ i.e. bijections $g:X\to X$ or $h:Y\to Y$ with the property:

$$g(\mu_X(x_1,\ldots x_n))=\mu_X(g(x_1),\ldots g(x_n))$$ $$h(\mu_Y(y_1,\ldots y_n))=\mu_Y(h(y_1),\ldots h(y_n))$$

How far can this be generalized? Even for "isomorphisms" between binary relations, that is bijective maps which preserve relations I found an analog of this theorem holds. That is the set of all isomorphisms between two relations can be expressed as any isomorphism between them composed with every automorphism on $X$ or $Y$ between them either left or right composed with said function. https://en.wikipedia.org/wiki/Isomorphism#Relation-preserving_isomorphism

I mean I don't really have a clear definition of "isomorphism" for general mathematical objects, just for algebraic structures, relations, and a few others. However it always seems this sort of theorem or an anlog of it holds. Does this fact/theorem have a name? Or is it just seen as obvious?

Answer 1

This is an artifact of category theory. The actual theorem it describes is rather trivial - the hard part is choosing the right definitions.

In particular, a category can be defined by a collection of objects with maps between them called morphisms. For instance, there is a category of sets where the objects are sets, and a morphism $X\rightarrow Y$ is simply a function $f:X\rightarrow Y$. There is a category of groups, whose objects are groups $(X,\cdot)$ and a map $(X,\cdot)\rightarrow (Y,*)$ is a homomorphism between these groups. You can similarly define a category of operations of a given arity (or, in universal algebra, of a given signature - which encompasses objects like rings).

More or less, a category is supposed to have a notion of composition - meaning that if I take a map $f:X\rightarrow Y$ and a map $g:Y\rightarrow Z$, then there is a map $g\circ f:X\rightarrow Z$. Functions and homomorphisms are closed under composition, so our examples are good. One requires that composition be associative and that, for every object $X$, there is an identity map $\operatorname{id}_X:X\rightarrow X$ which is an identity for composition.

Then, one defines an isomorphism as follows:

An isomorphism $f:X\rightarrow Y$ is a morphism such that there exists an inverse $g:Y\rightarrow X$. That is, a map such that $f\circ g = \operatorname{id}_Y$ and $g\circ f=\operatorname{id}_X$. This $g$ is known as $f^{-1}$.

So, in the category of sets, the isomorphisms are the bijections. In the category of groups, the isomorphisms are exactly what is usually called an isomorphism - and your extended notion of isomorphism also corresponds to a categorical one in the right category.

We also have the following definition:

An automorphism of an object $X$ is an isomorphism $f:X\rightarrow X$.

Then, your claim is essentially the following:

If $f:X\rightarrow Y$ and $g:X\rightarrow Y$ are an isomorphisms, then there is a $\sigma\in \operatorname{Aut}(X)$ such that $g=f\circ \sigma$ and a $\sigma'\in\operatorname{Aut}(Y)$ such that $g=\sigma'\circ f$.

This is easy to show after proving a few useful lemmas from the two given definitions: First, show that the composition of isomorphisms are isomorphisms. Second, show that the inverse of an isomorphism is an isomorphism. These both follow quickly from the axioms above. Then, let $\sigma = g\circ f^{-1}$ and $\sigma'=f^{-1}\circ g$. These are isomorphisms by the two lemmas I suggest, but are also maps $X\rightarrow X$ and $Y\rightarrow Y$ - thus automorphisms - by the properties of composition.