algebraic-topologygeneral-topology
Let $H$ denote the Hawaiian earring:
We defined a space $X$ to be semilocally simply connected if every point in $X$ has a nbhd. $U$ for which the homomorphism from the fundamental group of $U$ to the fundamental group of $X$, induced by the inclusion map, is trivial.
I'm looking for intuition on why $H$ is not semilocally simply connected.
The point is that topology the Hawaiian earring inherits from $\mathbb{R}^2$ is not the topology of the wedge sum of the circles which make it up. In particular, any open neighborhood of the origin in the Hawaiian earring completely contains all but finitely many of the circles, which is clearly not the case for an infinite bouquet of circles.
Hint:
Map the cross-links to ever smaller circles, using the fact that you can permute a finite number of circles in the earring homeomorphically.
(If you want the directions to match up, you'll need to cross each set of crosslinks in the middle, but the directions of each circle is not topologically distinguishable anyway, and it would make the figure more complex ...).
Best Answer
Consider any neighborhood of the point where the circles accumulate. At least one of the circles is completely contained in that neighborhood. A loop going around that circle is a non-trivial representative of the fundamental group of $U$ and it gets mapped by the inclusion to a non-trivial loop in $X$.
To show that these loops are non-trivial consider its image by the map $X\to S^1$ that collapses all but the circle in question to a single point. This image gives us the loop that goes around $S^1$ once, which is non-trivial in $\pi(S^1)$.