Given a set $X$. Let $S\subseteq X$ and consider $(X,S)$ as a very simple mathematical structure, lets call it a spotted set in analogy with pointed sets. Given two spotted sets, then a morphism $\alpha :(X,S)\longrightarrow(X^\prime,S^\prime)$ reasonably is a function $\alpha :X\longrightarrow X^\prime$ such that $x\in S\Rightarrow \alpha(x)\in S^\prime$.

In topology there is a spotted set $\tau\subseteq \mathcal{P}(X)$. Then morphisms are functions $\mathcal{P}(X)\overset{\alpha}{\longrightarrow}\mathcal{P}(X^\prime)$ such that $\mathcal{O}\in\tau \Rightarrow \alpha(\mathcal{O})\in \tau^\prime$. If there is a function $f:X^\prime\longrightarrow X$ such that $\alpha = \mathcal{Q}(f)$, where $\mathcal{Q}$ is the contra-variant power set functor, this correspond to Top and $f$ is continuous with respect to the topologies.

There are corresponding coincidences for several other structures, where the formulas of the morphisms can be derived, and my question is if there is an explanation to this correspondence?


Group-like structures as magmas and categories are characterized by relations $R\subseteq (X\times X)\times X$ and can obviously be expressed as spotted sets. Morphisms are functions $\alpha:(X\times X)\times X\longrightarrow(X^\prime\times X^\prime)\times X^\prime$ such that $((x,y),z)\in R \Rightarrow \alpha((x,y),z)\in R^\prime$. Functions $\alpha_1,\alpha_2,\alpha_3:X\longrightarrow X^\prime$ exists such that $\alpha((x,y),z)=((\alpha_1(x),\alpha_2(y)),\alpha_3(z))$ and if $\alpha$ is such that $\alpha_1=\alpha_2=\alpha_3$, then $\alpha_1$ correspond to group homomorphisms etc.

Action-like structures $R\subseteq (A\times X)\times X$. Here morphisms are functions $(A\times X)\times X\overset{\alpha}{\longrightarrow}(A\times X^\prime)\times X^\prime$ such that $((a,x),y)\in R \Rightarrow \alpha((a,x),y)\in R^\prime$. It exists functions $\alpha_0,\alpha_1,\alpha_2$ such that $\alpha((a,x),y)=((\alpha_0(a),\alpha_1(x)),\alpha_2(y))$. If $\alpha_0=1_A$ and $\alpha_1=\alpha_2$ this correspond to morphisms of actions.

Uniform spaces with a set of entourages $\phi\subseteq\mathcal{P}(X\times X)$. Morphisms are functions $\mathcal{P}(X\times X)\overset{\alpha}{\longrightarrow}\mathcal{P}(X^\prime\times X^\prime)$ such that $\mathcal{U}\in\phi \Rightarrow \alpha(\mathcal{U})\in \phi^\prime$. The condition on the morphisms of spotted sets to correspond to a uniformly continuous function is similar as above.

Multigraphs. Function $\varepsilon \subseteq E\times V^2$.

Undirected graphs. $E\subseteq\mathcal{P}(X)$, $e\in E\Rightarrow \alpha(e)\in E^\prime$, where $\alpha$ is a function $\mathcal{P}(X)\rightarrow\mathcal{P}(X^\prime)$.

It might be a good idea to point out that the formula for a morphism only depend on some outer structures. For example in magmas all formulas are the same and doesn't depend of "inner" conditions as associativity or inverses, with the exception of certain selected elements as a unit element.

If $\tau\subseteq \mathcal{P}(X)$ isn't a topology but is an other structure, the same formula would still be valid: $\alpha$ would be a morphism if there was a funcion $f:X^\prime\to X$ such that $\alpha = \mathcal{Q}(f)$ and $\mathcal{O}\in\tau\implies f^{-1}(\mathcal{O})\in \tau^\prime$.

There is an obvious an analogy with the Hom functor where magmas and universal algebras correspond to the covariant case, morphisms $\mathbb N\to X$, and the topological case correspond to the contravariant case with morphisms $X\to\mathbb N $.

  • 3
    $\begingroup$ I'm not sure in what sense this is a coincidence. It's not a very drastic reframing of the explicit definition of continuous functions. $\endgroup$ – Kevin Carlson Feb 9 '16 at 22:00
  • 3
    $\begingroup$ But I'm saying that the structures really aren't different. Yes, a topology is a special kind of spotted set. So what? It seems like you're basically making the observation that many objects are defined as structured sets, and that the structure on a set usually comes from a subset of a related set. $\endgroup$ – Kevin Carlson Feb 9 '16 at 22:54
  • 3
    $\begingroup$ It's just what a concrete category is. It's a fine point, but it seems unlikely that, for instance, you could prove theorems about spotted sets. $\endgroup$ – Kevin Carlson Feb 9 '16 at 23:59
  • 2
    $\begingroup$ Oh dear, the terminology... What you call a "spotted set" is almost universally called a pair of sets. Your category of "spotted sets" is, unsurprisingly, called the category of pairs of sets. I would strongly advise against calling your category $\mathsf{sSet}$, which is also almost universally the name of the category of simplicial sets, a completely different thing. $\endgroup$ – Najib Idrissi Feb 15 '16 at 10:37
  • 2
    $\begingroup$ It would be helpful to provide more examples of the "magic". I do see some magic here, but it has to do with the fact that a topology is not just a subset of $2^X$, but a sub-frame, and that $2^X$ is join-generated by atoms. But I have no idea whether this is related to other examples you have in mind. $\endgroup$ – Slade Feb 15 '16 at 11:20

If I understand correctly you call it a coincidence the fact that many concrete categories of structures can be encoded as some (sub-categories) of spotted sets (which by the way I would have called pair of sets, because they are very similar to pairs of spaces as studied in the framework of algebraic topology).

This does not come as a suprise to me and it is a consequence of the Bourbakian's belief that any mathematical structure can be encoded as a (family of) set(s) with operations and relations defined on them. Since every operation/function is traditionally encoded in set theories (such as ZFC) as a relation satisfying certain conditions, one could refine the above mentioned belief saying that every structure is a (family of) set(s) with relations defined on them.

Clearly if you regard structures as sets with relations (that is subsets of cartesian products, that is spotted sets) it becomes clear why all the structures can be embedded in the category of spotted sets.

Hope this answer you question.

Addendum: the following is just some additional material which could be skipped but I think it may be interesting because it somehow correlated to the subject considered (at least in my humble opinion).

You could also encode relations as operations (that is functions on sets): you could see any relation $R \subseteq A_1 \times A_2$ as a pair of functions $$(\pi_i \colon A \to A_i)_{i=1,2}$$ satisfying a condition, namely that the pair $(\pi_1,\pi_2)$ is jointly monoic.

In this way you can encode any relational structure as some sort of algebraic structure (over the family of sets $A_1,A_2$) and homomorphisms between such algebras correspond exactly to structure preserving morphisms.

This kind of algebraic construction is quite common, for instance it used to model many kind of dynamical systems such as finite state automata (and other kind of transition systems) as coalgebras in a opportune categories.

Basically the first kind of encoding (the one that uses relations) can be seen as the process of internalization of mathematical structures in the language of set theory while the second one (the one that uses operations) can be seen as the process of internalization of mathematical structures in the language of category theory.

  • $\begingroup$ The classical constructs are also subcategories of Set, but you can't derive the morphisms from that relationship. Interesting about the coding of relations as operators anyway... $\endgroup$ – Lehs Feb 16 '16 at 3:55

I see nothing "magical". A topological space is a set with certain additional properties. The category of topological spaces respects all properties of the category of sets.

  • $\begingroup$ Could you possibly elaborate on that? $\endgroup$ – Lehs Feb 9 '16 at 21:22
  • $\begingroup$ What do you mean when you wrote: "... The category of topological spaces respects all properties of the category of sets."? $\endgroup$ – Armando j18eos Feb 16 '16 at 15:50
  • $\begingroup$ "A topological space is a set with certain additional properties. " is simply not correct. It is a set together with structure. $\endgroup$ – HeinrichD Mar 22 '17 at 10:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.