Is there a mathematical structure without axioms that is equivalent to topological spaces?

Question

We can sometimes represent a "big structured" object by defining a "smaller unstructured" object. For example, a map $L: V \to W$ that satisfies the axioms of linear maps can be represented as a map $L: B_V \to W$, where $B_V$ is a basis for $V$. So there is a bijection between the space of linear maps $V \to W$ to the space of set-theoretic maps $B_V \to W$.

Is there similarly a family of mathematical structures without axioms, for which there is a bijection with the family of topological spaces?

Answer 1

I think the only reasonable general interpretation of what you decribe is that of a free object in a category. See also definition of a free object in a category.

In the category of vector spaces over a fixed field each object is free. In fact, each vector space is free on each its bases. In the category of abelian groups not every object is free. The free objects are denoted as free abelian groups.

What about the category $\mathbf{Top}$ of topological spaces and continuous maps? We shall show that the free objects are precisely the discrete spaces. This class of spaces is not very interesting and the answer to your question is "no".

For each space $X$ let $S(X)$ denote its underlying set and for each continuous map $F : X \to Y$ between spaces $X,Y$ let $S(F) : S(X) \to S(Y)$ denote its underlying function.

1. Let $X$ be discrete and $Y$ be an arbitrary space. Then each function $f : S(X) \to S(Y)$ is a continuous map $f : X \to Y$; that is, we have a bijection between the set of continuous maps $\mathbf{Top}(X,Y)$ from $X$ to $Y$ and the set of functions $\mathbf{Set}(S(X),S(Y))$ from $S(X)$ to $S(Y)$.

2. Let $X$ be a space which is free on some set $B$. This means that there exists a function $i : B \to S(X)$ such that for each space $Y$ and each function $f : B \to S(Y)$ there is a unique continuous map $F : X \to Y$ such that $S(F) \circ i = f$.
   Let us first verify that $i$ must be surjective. So assume $i(B) \subsetneqq S(X)$. Let $T$ be a two-point space with the trivial topology and $f : B \to S(T)$ be any non-surjective function. Since each function $F : X \to T$ is continuous, we conclude that there must be a unique function $\phi : S(X) \to S(Y)$ such that $\phi \circ i = f$. This is a contradiction because the condition $\phi \circ i = f$ allows that the elements $S(X) \setminus i(B)$ can be mapped arbitrarily into the two-point set $S(T)$, i.e. there exists more than one function satisfying our condition.
   Now let $Y$ be the discrete space with underlying set $S = S(X)$ and $f = i : B \to S(Y) = S$. Let $F : X \to Y$ be the unique continuous map such that $S(F) \circ i = i$. Then $S(F)$ is the identity on $S$ because $i$ is surjective. Hence the identity $X \to Y$ is continuous which implies that $X$ is discrete.

Answer 2

I'm not sure if this is close enough to your question, but it's something vaguely related that came to mind.

An interesting idea in algebraic geometry is that going between algebra and geometry is via opposite categories. (They take this correspondence very literally -- "algebra" is the category of commutative rings $\mathrm{CRing}$, and "geometry" is just $\mathrm{CRing}^{\mathrm{op}}$, objects of which are known as affine schemes.)

We might try to do this in reverse: $\mathrm{Top}^{\mathrm{op}}$ ought to be an "algebraic" description of topological spaces. I'm not familiar with the full answer (it appears there is a description of this category), but I do know that restricting to certain subcategories of $\mathrm{Top}$ gives some nice answers. For example, the category of compact totally disconnected Hausdorff spaces (the profinite spaces) is equivalent to the category of boolean algebras. This is an instance of Stone duality.