Let me first give the definition of a local coefficient system (see also [2, p. 257] and [3, p. 35]):

Let $X$ be a topological space. A local coefficient system is a functor from the category $\Pi_1(X)$ (= the fundamental groupoid) to the category AbGrp of abelian groups. Such a functor assigns to each $x \in X$ an abelian group $G(x)$ and to each homotopy class $\xi \in \pi_1(X; x_1, x_2)$ (the set of all endpoint-preserving homotopy classes of paths from $x_1$ to $x_2$) a homomorphism $G(\xi): G(x_2) \to G(x_1)$; these are required to satisfy

(i) if $\xi \in \pi_1(X, x) = \pi_1(X; x, x)$ is the identity, then $G(\xi): G(x) \to G(x)$ is the identity;

(ii) if $\xi \in \pi_1(X; x_1, x_2)$, $\eta \in \pi_1(X; x_2, x_3)$, then $G(\xi \eta) = G(\xi) \circ G(\eta): G(x_3) \to G(x_1)$

So my question is:

It is easy to see that every bundle of groups (defined in [1, p. 330]) is a local coefficient system, but I think the converse is not true (as stated without proof in [3, p. 35]), so I am looking for a (non-trivial) local coefficient system which is not a bundle of groups.

Please note: A good choice for the space $X$ in the above definition of a local coefficient system could for example be the Hawaiian earrings.

Thank you in advance!


[1] Hatcher, Allen: Algebraic Topology, 2002, https://www.math.cornell.edu/~hatcher/AT/AT.pdf, p. 330

[2] Whitehead, George: Elements of Homotopy Theory, New York: Springer, 1978 (Graduate Texts in Mathematics Vol. 61), p. 257

[3] Hutchings, Michael: Introduction to higher homotopy groups and obstruction theory, 2011, https://math.berkeley.edu/~hutching/teach/215b-2011/homotopy.pdf, p. 35

  • 3
    $\begingroup$ The converse is true for spaces that have a universal cover, so counterexamples will be quite pathological. $\endgroup$ – Andy Putman Aug 1 '16 at 20:01
  • 1
    $\begingroup$ An alternative definition is that a local system is a locally constant sheaf of abelian groups on $X$. This agrees with your definition for nice $X$ but I think not in general. $\endgroup$ – Qiaochu Yuan Aug 1 '16 at 23:31

Strip away the group structure and you get the simpler question: Does every functor from the fundamental groupoid of $X$ to Set correspond to a covering space (a bundle of sets)? As a special case this includes the question, does every subgroup of the fundamental group of a path-connected space come from a connected covering space.The answer to all of these is no. A standard example is when $X$ is the "Hawaiian earring" a.k.a. "clamshell space" a.k.a. "shrinking wedge of circles" (union of infinite sequence of circles in the plane with diameters tending to zero and one point in common).

I like to organize these ideas like this: There is a functor, obviously faithful, from the category $Cov(X)$ of covering spaces of $X$ to the category $Fun(\Pi_1(X),Set)$. It is not always fully faithful, but it is if $X$ is locally path-connected. In this case it is not always an equivalence of categories, but it is if $X$ is, as they say, "semi-locally simply connected".

  • $\begingroup$ Thank you very much for your quick answer and the given intuition! Unfortunately, I was not yet able to explicitly write down a local coefficient system for the Hawaiian earrings which is not a group bundle. A reference where I can find such a local coefficient system would also be helpful. $\endgroup$ – Kathy Aug 2 '16 at 19:04
  • 2
    $\begingroup$ You should check that a necessary condition for a local coefficient system $G$ to be a bundle of groups is that every point in $X$ has a neighborhood $U$ such that for all loops $\xi \in \pi_1(U,x)$ the map $G(\xi)$ is the identity of $G(x)$. $\endgroup$ – user95545 Aug 3 '16 at 11:51
  • 2
    $\begingroup$ One can write down an example with the Hawaiian earrings $X$ where the fiber at the special point is $G(x)=\prod_{j=1}^\infty \mathbb Z$, and for each loop $\xi\in \pi_1(X,x)$, the map $G(\xi)$ multiplies the $j$-th component of $G(x)$ with $(-1)^{w_j}$, where $w_j$ is the winding number of $\xi$ around the $j$-th circle. $\endgroup$ – user95545 Aug 3 '16 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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