Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am thinking about a "correct" didactic way of linking topology (algebraic topology) to geometry. Usually, we are taught introducing geometry first, then topology, almost as an abstraction of geometry.

So how can we link back?

For example, when we have a chain complex, how can we introduce formally an affine space and have this complex endowed with geometric features?

Cheers & Thanks!

share|improve this question
"Abstraction of geometry": isn't it a good description of "topology"? –  Did Feb 4 '13 at 14:43
Given a chain complex of finite rank abelian groups, you can build a CW complex that has this as its cellular homology. Just let generators of your groups correspond to cells, and attach them via maps that come from the boundary operator in the chain complex. I'm not sure if this is what you are after. –  Grumpy Parsnip Feb 4 '13 at 14:45
Following up on Did's comment, you can make an argument that point-set topology is somewhat inspired by analysis, trying to generalize the notion of "convergence". –  Isaac Solomon Feb 4 '13 at 15:06
Yes @Did, I usually introduce topology as an abstraction of geometry, but in everything that I introduce I never look back. –  senseiwa Feb 4 '13 at 15:13
@JimConant you suggest to map C_p(G) (sorry, I don't know how to type mathematically in commments) to, say, a simplicial complex? If so, shouldn't it suffice to introduce a map between C_0(G) to K_0? –  senseiwa Feb 4 '13 at 15:17

1 Answer 1

up vote 3 down vote accepted

Chain complex happen to be the simplicial objects in the category of modules. To any simplicial object, it is possible to construct funtorially a topological space, its geometric realization.

This functor has the very nice property to be a left ajoint to the singular functor $S$ which associates to any topological space, its singular chain complex.

$$ \mathrm{Hom}_{\mathrm{Top}} (|X|, Y) = \mathrm{Hom}_{\mathrm{Ch}}(X, SY)$$

share|improve this answer
This is interesting, I didn't think about using category theory, but it works. What book do you suggest I should take a look at? (I usually suggest reading Hatcher and Munkres) –  senseiwa Feb 4 '13 at 15:20
Appendix of Hatcher's book. –  Damien L Feb 4 '13 at 15:24
Thanks, it's a good thing to remind that appendices matter, then! PS. Why isn't the site auto-completing your name, when up above it does? –  senseiwa Feb 6 '13 at 13:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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