Semi-local simple connectedness is a property that arises in Algebraic Topology in the study of covering spaces, namely, it is a necessary condition for the existence of the universal cover of a topological space X. It means that every point $x \in X$ has a neighborhood $N$ such that every loop in $N$ is nullhomotopic in $X$ (not necessarily through a homotopy of loops in $N$). The way I see it, the prefix "semi-" is refers more to "simply connected" than to "locally", since if such $N$ exists, all other neighborhoods of $x$ inside $N$ also have the property, so each point has a fundamental system of (open) neighborhoods for which the property holds (EDIT see Qiaochu's comment). Instead it isn't true that a semi-local simply connected space is locally simply connected (i.e each point has a fundamental system of open, simply connected neighborhoods): take the space $$ X = \frac{H \times I}{ \sim } $$ where $H$ is the "Hawaiian earring", or infinite earring (which is an example of non semi-locally simply connected space) and $\sim$ is the equivalence that identifies $H \times \{0\}$ to one point.

However I was interested in finding another type of counterexample. Consider the topological property (call it $*$) consisting in the existence, for all $x \in X$, of a simply connected, not necessarily open, neighborhood of $x$. We have $$ \text{semi-local simple connectedness} \Leftarrow * \Leftarrow \text{local simple connectedness} \wedge \text{simple connectedness} $$ I am wondering if $$ \text{semi-local simple connectedness} \Rightarrow * $$ holds. Intuitively it shouldn't, but I'm having trouble finding a counterexample. For example, the space $X$ described above won't work because it is simply connected (even contractible). It seems to me that, if a counterexample does exist, it must have local pathologies (to ensure that a certain point $x$ doesn't have a simply connected neighborhood), and globally the space should allow loops close to $x$ to be nullhomotopic, but in such a way that every neighborhood $N$ of $x$ contains a small enough loop that will not contract in $N$. EDIT Also, I am looking for a counterexample which is a locally path connected space (a previous answer showed a counterexample without this property.. but @answerer it was interesting anyway, you shouldn't have deleted it!)

But perhaps I'm wrong, and the two assertions are equivalent, or maybe I am missing something very simple. If anybody has any ideas please share, thank you!

  • $\begingroup$ If you generally interested in counterexamples for point-set topology you might like this book amazon.co.uk/Counterexamples-Topology-Dover-Books-Mathematics/… The math library in my old uni actually had it available. $\endgroup$ Jul 19, 2012 at 14:03
  • $\begingroup$ @SimonMarkett Thanks, my library has it too and it's the first place I checked.. but unfortunately there's nothing on semi-local simple connectedness. $\endgroup$ Jul 19, 2012 at 14:10
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    $\begingroup$ The "semi" modifies "locally." It refers to the fact that the homotopy doesn't have to stay in $N$. $\endgroup$ Jul 19, 2012 at 15:12
  • $\begingroup$ @QiaochuYuan what I meant with that remark is that (quoting from Hatcher - Algebraic Topology, page 62) "When we say a space has a certain property locally, [...], we shall mean that each point has arbitrarily small neighborhooods with this property". But it is true that in this case the property that we want arbitrarily small neighborhoods (or just one neighborhood, it's equivalent) to satisfy has a global nature. $\endgroup$ Jul 19, 2012 at 15:30
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    $\begingroup$ @Emilio: yes, but the existence of that homotopy isn't a property of the neighborhood (as a space in and of itself) but a property of how the neighborhood sits in the larger space. $\endgroup$ Jul 19, 2012 at 15:31

1 Answer 1


Call the "interesting" point in the Hawaiian earring $q$ (the point where all the circles intersect). In your space $X$, glue the quotiented point $\{H \times \{0\}\}$ to the point $(q,1) \in H \times I$, essentially bending your cone around in a loop. Call the resulting quotient space $Y$, and the glued point $p$. Note that $p$ could be described either as $(q,1)$ or as $\{H \times \{0\}\}$. ($Y$ can also be described as the Mapping Torus of the map that takes the entire Hawaiian earring to the point $q$).

$Y$ is semi-locally simply connected but doesn't satisfy property $∗$.

To see $Y$ is semi-locally simply connected, note that any point has a neighborhood that is disjoint from a set of the form $H \times \{x\}$ for some $x \in I$. The inclusion of such a neighborhood into $Y$ is nullhomotopic (simply retract the portion contained in $H \times [0,x)$ to $p$, then follow the strong deformation retraction of $X$ to $\{H \times \{0\}\}$), which certainly implies that any loop contained in it is nullhomotopic in $Y$.

But $p$ itself has no simply-connected neighborhood. Any such neighborhood $N$ contains a loop $L \times \{1\}$ in $H \times \{1\}$, which could only be contracted in $N$ by including all of $L \times I$ in $N$. In particular, you'd have to include all of $\{q\} \times I$ in $N$, But $\{q\} \times I$ is itself a loop which isn't nullhomotopic in $Y$, and so also isn't nullhomotopic in $N$. Thus, $N$ cannot be simply connected.

  • $\begingroup$ +1, this is a very nice idea! I don't know whether I really can follow your argument for why it works, though. Maybe it would help, if you could make it just a bit more precise and formal? Anyways, I think I can understand what you mean and I'm pretty sure this works. $\endgroup$
    – Sam
    Jul 20, 2012 at 1:41
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    $\begingroup$ @SamL. I've fleshed out the argument a bit. Does it make more sense to you now? Basically to contract a small circle in $H \times \{1\}$, you have to go all the way around a loop that isn't contractible in the full space. $\endgroup$ Jul 20, 2012 at 4:24
  • $\begingroup$ Yes, I think this is much clearer now. Thank you! I can't +1 twice, unfortunately. $\endgroup$
    – Sam
    Jul 20, 2012 at 10:04
  • $\begingroup$ @MartianInvader Yes, this works nicely! Thanks a lot for finding this counterexample and explaining it very clearly. $\endgroup$ Jul 20, 2012 at 11:18

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