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I came across this example Spanier's Algebraic Topology book (by way of the Munkres Topology book). I kind of have an intuitive idea of why the space isn't simply connected but can't figure out a formal way to prove it isn't.

The construction of the space is as follows:

Let $C_n$ be the circle of radius $1/n$ centered at $(1/n , 0)$ and take $X$ to be the union of all the spaces $C_n$. If we imbed $\mathbb{R}^2$ in $\mathbb{R}^3$ as points where $z = 0$ then we take $Y$ to be the closed lines from $(0,0,1)$ to $X$. We will take $Y'$ to be the reflection of $Y$ through the origin. $Y \cap Y'$ is a single point and both $Y$ and $Y'$ are simply connected but their union is not.

I can defintely see why both $Y$ and $Y'$ are simply connected because they can be retracted to the point at the top of the cones and I can sort of guess that the reason $Y \cup Y'$ isn't simply connected is that you have these two separate points and you can't contract to both of them all at once and part of the issues that keeps you from doing this is how there are infinitely many circles on each side (and if it was only finitely many then we would have a simply connected space).

The ways that I have thought about solving this mostly have been finding a homotopy equivalence to a familiar space where I know the fundamental group already or finding suitable open set $U$ and $V$ so I can use the Seifert van Kampen theorem.

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  • $\begingroup$ I'm slightly confused. $Y$ is the closed line segments from $(0,0,1)$ to $X$? Is $Y'$ the reflection of $Y$ across the plane $z=0$? I don't think reflection about a point is well defined. $\endgroup$ – Seth Dec 14 '14 at 23:47
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    $\begingroup$ My suggestion doesn't agree with $Y\cap Y'$ is a single point, so it must not be what you meant. Please clarify. $\endgroup$ – Seth Dec 14 '14 at 23:49
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    $\begingroup$ @Seth I was using the term from the book. How I was picturing it is like this wildtopology.files.wordpress.com/2014/06/griffiths1.png $\endgroup$ – user171177 Dec 14 '14 at 23:51
  • $\begingroup$ Maybe each point is reflected across the plane through the origin perpendicular to the line from the point to the origin? $\endgroup$ – Seth Dec 14 '14 at 23:51
  • $\begingroup$ That looks like it might be the right way to describe the process (I really did pull that description straight from the book so I didn't decide on that word myself) $\endgroup$ – user171177 Dec 14 '14 at 23:52
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After posting this question I came across this blog post http://wildtopology.wordpress.com/2014/06/28/the-griffiths-twin-cone/ and now feel like I can answer my own question.

As mentioned in the comments the space I describe looks like this Griffith Cone

but it is homeomorphic to a space that looks like this Griffith Cone

Using the second of these two spaces and two applications of the Seifert–van Kampen theorem first with $U$ and $V_\text{odd}$ and then with $U \cup V_\text{odd}$ and $V_{\text{even}}$ we can see that the inclusion map identifying the Hawaiian earring with the intersection of the space with the $xy$ plane induces a surjection of fundamental groups.

The kernel of this map is the conjugate closure of the union of subgroups that contain loops only going around either even or odd circles but not both. This is not the entire fundamental group because the loop that goes once around every circle from biggest to smallest is not contained in this union.

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