MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $X$ is a path connected space such that every connected covering space of $X$ is trivial (1-fold.) Must $X$ be simply connected?

Intuitively, the answer seems to be no (imagine taking a disk, cutting out a square, and gluing in $T\times T$ where $T$ is the topologist's sine curve.) But this is a rather weak intuition.

share|cite|improve this question
Nitpick: I assume you mean "connected covering space", because otherwise... – Chad Groft Mar 13 '11 at 16:10
Yes, thank you. – David Cohen Mar 13 '11 at 16:14
up vote 13 down vote accepted

No, the harmonic archipelago (An illustration is on pg 7 of W. A. Bogley and A. J. Sieradski, Universal path spaces, preprint) is a locally path connected subspace of $\mathbb{R}^{3}$ and has uncountable fundamental group but every connected cover is trivial.

This is part of a more general phenomenon. Let $\pi_{1}^{top}(X)$ be the fundamental group of a space $X$ with the quotient topology of the loop space $\Omega(X)$ with the compact-open topology (sometimes called the "topological fundamental group"). This is a quasitopological group (in that inversion is continuous and multiplication is continuous in each variable) but is not always a topological group. If $p:X\rightarrow Y$ is a covering map the induced homomorphism $p_{\ast}:\pi_{1}^{top}(X)\rightarrow \pi_{1}^{top}(Y)$ is an open embedding of quasitopological groups (i.e. $\pi_{1}^{top}(X)$ embeds as an open subgroup). One consequence of this is that if $\pi_{1}^{top}(X)$ has the indiscrete topology, then either $\pi_{1}(X)=1$ or every connected covering of $X$ is trivial. There are lots of examples of spaces where this occurs other than the harmonic archipelago.

share|cite|improve this answer
See also Jeremy Brazas. Semicoverings: a generalization of covering space theory, Homology, homotopy and applications, 14(1), (2012) 33-63. – Ronnie Brown Apr 29 '12 at 16:20

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.