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.



Is a space with no covering spaces simply connected?
No, the [harmonic archipelago](https://arxiv.org/abs/math/0501426) (An illustration is on pg 7 of [W. A. Bogley and A. J. Sieradski, Universal path spaces, preprint](https://web.archive.org/web/20210506162111/http://people.oregonstate.edu/%7Ebogleyw/research/ups.pdf)) 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.


