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Let’s consider the following pseudo-definition of (nice) topological spaces : a space is a set $X$ together with distinguished paths $[0,1]\to{}X$ called continuous paths, distinguished maps $[0,1]\times[0,1]\to{}X$ called homotopies, and so on in every dimension (in a globular style), satisfying a bunch of properties. For example, the constant path should be continuous, the composition of two continuous paths is continuous, the slices of an homotopy are continuous paths, etc.

Is there a way to formalize precisely this definition, and how?

I apologize for the vagueness of the question.
And this is just a curiosity, I had to talk about topology to computer scientists a few days ago and the most intuitive definition of topological space I found is to say that a topological space is a set with a well-behaved notion of continuous path. And now I’m wondering whether this could be an honest definition of topological spaces.

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  • $\begingroup$ Do you mean "joining of two continuous paths is continuous"? $\hspace{2 in}$ Composition of paths would in general not make sense. $\;\;$ $\endgroup$
    – user5810
    Jan 27, 2012 at 4:13
  • $\begingroup$ I think concatenation of continuous paths is commonly called "composition" by category theorists and homotopy theorists, thinking of paths as like morphisms in a category of points. $\endgroup$ Jan 27, 2012 at 5:49
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    $\begingroup$ Strange things can be done with this concept. I cannot find the reference in Math Reviews, but I recall a paper where the author put a topology on the plane so that the only continuous paths were piecewise linear. $\endgroup$
    – Matt Brin
    Jan 27, 2012 at 6:32
  • $\begingroup$ @Ricky Yes, composition is concatenation. $\endgroup$ Jan 27, 2012 at 12:12
  • $\begingroup$ @GuillaumeBrunerie you're asking if knowing what are the continuous functions of type $I^n \to X$ characterize the topology, am I right? $\endgroup$ Jan 27, 2012 at 15:38

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I don't think I've seen a definition of a space-like notion phrased only in terms of paths, but you could certainly write one down, perhaps as an example of concrete sheaves. It seems related to the notion of Froelicher space which defines "smoothness" in terms of paths and "co-paths."

However, it seems unlikely to me that you'll get a very good notion without including also some higher-dimensional test objects in addition to intervals. For instance, there is the notion of $\Delta$-generated space, which can be described as a set $X$ with sets of distinguished maps $\Delta^n \to X$ for all $n$, satisfying a bunch of properties. I don't know any description of those properties other than "if $X$ is given the final topology induced by the distingished maps, then all continuous maps $\Delta^n \to X$ are distinguished", but that doesn't mean there isn't one; I don't know whether anyone has looked.

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  • $\begingroup$ Sorry if it wasn’t clear enough, but I’m not only including the paths, but also the homotopies between paths, homotopies between homotopies between paths, and so on. This seems indeed very similar to the notion of $\Delta$-generated space, thanks (but another descriptions of those properties would be nice) $\endgroup$ Jan 27, 2012 at 17:00
  • $\begingroup$ Something like a cube-generated space? $\endgroup$
    – David Roberts
    Jan 30, 2012 at 22:12
  • $\begingroup$ Well, cubes are homeomorphic to simplices (both are homeomorphic to balls), so a cube-generated space would be the same as a $\Delta$-generated one. $\endgroup$ Feb 1, 2012 at 0:18
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    $\begingroup$ The existence of space filling curves means that $\Delta$-generated spaces and interval generated spaces are the same. $\endgroup$
    – Jeff Smith
    Feb 13, 2012 at 1:14
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Try the wikipedia entry on sequential spaces. I quote:

"Many conditions have been shown to be equivalent to $X$ being sequential. Here are a few:

$X$ is the quotient of a first countable space.

$X$ is the quotient of a metric space.

For every topological space $Y$ and every map $f : X \to Y$, we have that f is continuous if and only if for every sequence of points (xn) in X converging to x, we have (f(xn)) converging to f(x).

The final equivalent condition shows that the class of sequential spaces consists precisely of those spaces whose topological structure is determined by convergent sequences in the space."

The point is that given a set of test maps from $I$ to a set $X$ you can put a topology on $X$ as the quotient of the disjoint union of copies of $I$ one for each test map. And then $X$ will be the quotient of a first countable space.

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No such definition can be given for a general topological space, because a general topological space can have very few morphisms from or into $[0;1]$. E.g. consider $\Bbb Q$: there are no non-trivial continuous paths, containing only rational points. On the other side, Urysohn's lemma states that only normal topological spaces have enough continuous functions on them.

Your proposed axiomatization looks very similar to the axioms of $(\infty,1)$-categories. There you have objects, corresponding to points, morphisms between them, corresponding to paths, 2-morphisms between morphisms, corresponding to homotopies between paths etc. Every topological space has a canonically associated $(\infty,1)$-category, constructed just as above. There is an adjunction between the bicategories of topological spaces and of $(\infty,1)$-categories. In fact, it is even a Quillen equivalence (equivalence of homotopy categories) for some natural choices of model structure on categories (i.e. for some notion of homotopy equivalence). However, the image of this adjunction in $\mathcal{T}op$ consists not of all spaces, but only of CW-complexes. So the answer is: it works, but only for very nice spaces.

You can look further information on $(\infty,1)$-categories and quasicategories. A short survey of theory can be found in the first chapter of J. Lurie's "Higher topos theory".

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