## Algebraic topology of finite topological spaces August 11, 2007

Posted by Noah Snyder in Algebraic Topology, fun problems.

Here’s a fun question that was floating around Mathcamp last week: find a finite topological space which has a nontrivial fundamental group. One answer to this question after the jump.

One example is a space S with 4 points, two of which are open and two of which are closed. First, consider the line with the origin doubled. Now quotient out by setting all positive points equal to each other, and all negative points equal to each other. This gives a four point space S.

There’s a map from the circle to S given by sending your favorite two points on the circle to the closed points, and the two open intervals between them to the open points. It is not difficult to see that this cannot be extended to the disc. A better proof is to exhibit S’ the universal cover of S. The space S’ looks like:

The points in the middle column are closed. The points in the other two columns are open, and the closure of any such point contains the two nearest points in the middle column. S’ is not contractable, but any compact (i.e. finite) subset of it is contractable, so it is simply connected. Hence $\pi_1(S) \cong \mathbb{Z}$ since the deck transformations of S’ just come from shifting up and down.

Here are two more fun problems: find all the homology and homotopy groups of this 4 point space.

1. James - August 11, 2007

For general topological spaces, you wouldn’t expect the usual fundamental group defined in terms of paths to still classify covering spaces. But I think I remember hearing that there is still a Grothendieck-style definition of a fundamental group that classifies finite-degree covering spaces. In the special case of a CW complex, this would be the profinite completion of the usual fundamental group. I don’t know if possible to do it without the finite-degree restriction. Maybe it’s just what comes out of Grothendieck’s formalism, which was creating with algebraic fundamental groups in mind.

2. Noah Snyder - August 12, 2007

Acording to Wikipedia (and to Eric, a camper and our resident expert on point-set topology) a space has a universal cover if and only if it is path-connected, locally path-connected, and semi-locally simply connected. All of these conditions are easy to check for S.

3. Ben Webster - August 12, 2007

James-

There’s a more categorical way of thinking about the fundamental group: if you have any reasonable notion of “covering space,” then one can take the category of covering spaces of a given one. Call this $\mathcal C$.

For any reasonable notion of covering space (in particular, the usual topological one), one has a functor $\mathcal{C}\to \mathsf{Set}$ sending a cover to the fiber over a generic point. This functor is even monoidal for the “tensor product” on coverings given by fiber product. By analogy with the Tannakian formalism, one can define the fundamental group of $X$ for a given notion of covering to be the automorphism group of this forgetful functor.

If $X$ has a universal cover, then you can check that you’ll get back the usual fundamental group. If you restrict to finite covers, you’ll get the profinite completion of the fundamental group. If you switch to the algebraic category, you should get Grothendieck’s algebraic fundamental group. The reason that you get a profinite group here is that the algebraic restriction forces you to only consider finite covers.

4. James - August 13, 2007

Right. What I meant was that I remember hearing that you always have a Galois category (i.e. the finite discrete version of a Tannakian category) for any topological space whatsoever. And so even though you can’t always define pi_1, you can always define its completion, or rather what would be its completion if pi_1 existed.

5. carnahan - August 20, 2007

It looks like any sufficiently subdivided CW complex can be rendered as a locally finite topological space in the same way as you did with the circle. In particular, you should be able to get any finitely presented group as $\pi_1$ of a finite topological space.

Are there interesting questions about finite “homotopy types”? It’s not clear that this adds anything new to algebraic topology.

Incidentally, there are multiple algebraic categories (e.g., tame, etale, Nisnevich), coming from different notions of cover, and they yield very different fundamental groups.

6. Todd Trimble - August 20, 2007

Cool “postcards” from mathcamp, Noah! Your entry here got me thinking:

There is an equivalence of categories

O: FinTop –> FinPreOrd

between finite topological spaces and finite preorders,
where the order –> in O(X) is defined by x –> y iff x is
contained in the closure of y. For Noah’s 4-point example S, the associated preorder O(S) looks like

a d
a d

with b and c both pointing to a and to d (no other relations).

On the other hand, one can take the classifying space of a finite preorder

B: FinPreOrd –> Top

as usual, by taking geometric realization of the nerve of the preorder (considered as a category). On Noah’s example S, the classifying space of the associated preorder, BO(S), is a circle S^1.

The map S^1 –> S that Noah defined generalizes: for finite topological spaces X, I believe I can define a continuous map

BO(X) –> X,

almost as a piece of pure category theory. In the end, it comes down to defining a continuous map

Aff(n) –> D(n)

from the n-dimensional affine simplex to the finite topological space with n+1 points represented by the preorder Delta_n = (0 –> 1 –> … –> n). I’ll leave this to the imagination for now (details available on request).

Then, does anyone know what can be said of this map BO(X) –> X in terms of homotopy? For example, does pi_1 induce an isomorphism? What happens with higher homotopy groups?

7. Eric - August 21, 2007

If X is T_0 (I haven’t checked whether it still works for non-T_0 spaces, the map BO(X) -> X (which is a quotient map) turns out to have a nice universal property: any map Y -> X lifts to BO(X), as long as Y is sufficiently nice (metrizable or a CW complex, say; the actual condition is hereditary perfect normality). Furthermore, the lift is unique up to a homotopy such that every stage of the homotopy is a lift. It’s easy to see that this implies that the map induces isomorphisms on all homotopy groups. You can either use this to show it also induces isomorphisms on homology, or you can prove that directly by induction on the number of points and Mayer-Vietoris.

Any barycentric subdivision of a simplicial complex C is BO(X), where X is the poset of faces of C ordered by inclusion. Thus every finite simplicial complex has a finite “model”.

8. Todd Trimble - August 21, 2007

Thanks, Eric — very useful reply. I think the weak homotopy equivalence for finite T_0 spaces implies the same holds for all finite spaces:

A finite space X is T_0 iff its associated preorder is a poset, and every preorder P is equivalent as a category to a (unique up to isomorphism) poset P’, with P’ a retract of P. It’s well known that the categorical equivalence implies BP and BP’ are homotopy equivalent. On the other hand, the equivalence P ~ P’ means there is a preorder map

(0 –> 1) = 2 –> hom(P, P)

sending 1 to the identity and 0 to a factoring through P’. Now switch to the topological picture, and pull back along the evident continuous map I = [0, 1] –> 2 to conclude that P and P’ are homotopy equivalent as spaces.

It now follows from naturality of BO(X) –> X that this map is a weak homotopy equivalence for all finite X.

9. Benjamin Steinberg - June 1, 2012

This is way late, but McCord showed any finite simplicial complex is weakly equivalent to its poset of faces with the Alexandrov topology so you can get any finitely presented group.

In particular the nerve of a poset is weakly equivalent to the poset.

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