# Is a space with no covering spaces simply connected?

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.

• Nitpick: I assume you mean "connected covering space", because otherwise... – Chad Groft Mar 13 '11 at 16:10

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.