In Algebraic topology we use tools from abstract algebra: we ask question like "when are two spaces not homeomorphic?" by associating algebraic objects to them.

For example, the fundamental group of a topological space.

Let's go one step further and increase the number of binary operations from one to two, such as with rings. So I am curious whether there are other algebraic objects with two binary operations associated to topological spaces. Basically I am curious to know whether rings and fields are also associated to topological spaces, or is it just a vague question to ask?


4 Answers 4


Yes, there are such structures, e.g. a cohomology ring:


Actually they are quite important. For example by analysing the cohomology ring of spheres it can be deduced that there is no topological group structure on spheres except for dimensions $0, 1, 3$ and somewhat weaker (i.e. not associative) topological group structure in dimension $7$.

Actually in case of topological groups this goes even further. The cohomology ring (of a topological group) becomes a Hopf algebra which is a very rich algebraic structure: it's a vector space with multiplication and comultiplication. So it has at least four operators ($+$, scalar multiplication, vector multiplication, comultiplication).

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    $\begingroup$ Not to mention the homology coring. :P $\endgroup$ Oct 26, 2016 at 21:04

Algebraic topology is broadly speaking the study of functors (and natural transformations between them) from the category of topological spaces (or related categories such as its homotopy category) to an "algebraic" category. You get a much better perspective on the methods and "big picture" in algebraic topology when you adopt this point of view.

Here's a nice but easy example: for $R$ a topological commutative ring, the assignment of the $R$-algebra of continuous $R$-valued functions on a topological space is a (contravariant) functor from topological spaces to $R$-algebras. In fact, when the target ring is the complex numbers, this functor induces an adjoint equivalence of categories between compact Hausdorff spaces and (the opposite of) the category of commutative C* algebras. This is called Gelfand duality.

There's plenty of other interesting functors to algebraic categories. For instance, the singular cohomology ring gives you a functor from topological spaces to rings. The singular chain complex functor is a functor to chain complexes of abelian groups. The fundamental groupoid is a functor from topological spaces to groupoids.

Since I mentioned that natural transformations between functors are also of interest in algebraic topology, here's a key example: characteristic classes. Let $\mathsf{hoTop}$ denote the category of topological spaces with continuous maps modulo homotopy equivalence as the arrows. Since pullbacks of principal $G$-bundles by homotopic maps are isomorphic (at least when we restrict $\mathsf{Top}$ to be the category of "nice spaces" such as CW complexes and manifolds), then we have a (contravariant) functor: $$B_G: \mathsf{hoTop}^{\mathrm{op}} \to \mathsf{Sets}$$ that assigns to each space $X$ the set of isomorphism classes of principal $G$-bundles over $X$. Now given some cohomology functor $H$, lets compose it with a forgetful functor from abelian groups (or rings or whatever) to $\mathsf{Sets}$. Then we can think of $H$ as a (contravariant) functor $$H: \mathsf{hoTop}^{\mathrm{op}} \to \mathsf{Sets}.$$ Since these two functors have the same source and target, we can talk about natural transformations between them. In fact natural transformations $c: B_G \Rightarrow H$ are precisely characteristic classes!

Lastly, here's a "stupid but correct" answer to your question: pick any sort of category of algebraic objects, as wild as you like, and let's call it $\mathsf{A}$. Now pick your favourite algebraic object $A$ in this category. Now construct a functor $F: \mathsf{Top} \to \mathsf{A}$ that maps every topological space $X$ to the algebraic object $A$, and takes every continuous map $f: X \to Y$ to the identity map $\mathrm{id}_A: A \to A$ on $A$. Then this is a perfectly reasonable (but rather useless) construction in algebraic topology.


The example of a cohomology ring was already given, so I'll mention the K-theory ring. https://en.m.wikipedia.org/wiki/K-theory

Topological K-theory studies the Grothendieck group of homotopy classes of vector bundles on a space where the operation is direct sum. The ring multiplication is given by the tensor product. This is a commutative ring.


Well, there is a "stupid" example:

The set of opens is a complete Heyting algebra.

  • $\begingroup$ by open you mean open sets ? $\endgroup$ Oct 26, 2016 at 21:03
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    $\begingroup$ Yes, the open sets ordered by inclusion (i.e. the topology). $\endgroup$ Oct 26, 2016 at 21:05

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