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Let $H$ be the Hawaiian earring and let $H'$ be the reflection of the Hawaiian earring across the $y$-axis (in the Wikipedia picture). There is a canonical homomorphism from the free product $\pi_1(H) * \pi_1(H')$ to $\pi_1(H \cup H')$ (with basepoint their intersection), but it is not an isomorphism.

This was intended to be a recent homework problem of mine, but as stated the problem actually asked whether the two groups are abstractly isomorphic. I don't know the answer to this question, and neither does my professor. My guess is that they are not isomorphic, but I don't have good intuitions about such large groups.

Edit: to be clear, I know how to do the intended problem, and I also know that $H \cup H'$ is homeomorphic to $H$.

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@Qiaochu Yuan: So the question is whether $\pi_{1}(H)$ is isomorphic to $\pi_{1}(H')$? –  PEV Feb 19 '11 at 0:54
    
@PEV: no. The question is whether $\pi_1(H) * \pi_1(H)$ is isomorphic to $\pi_1(H \cup H')$. –  Qiaochu Yuan Feb 19 '11 at 1:00
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I had a conversation about it with Jim Conant here: math.stackexchange.com/questions/21705/… I think there may be something about it in the paper "The second van-Kampen theorem for topological spaces" of Andreas Zastrow, but I can't download it from home... –  Max Feb 19 '11 at 1:32
    
My best guess is that Cannon and Conner's presentation of the Hawaiian earring group in terms of transfinite words can be used to show the groups are not isomorphic. However, I don't actually see how to do that. –  Grumpy Parsnip Feb 19 '11 at 1:55
    
@Qiaochu Yuan: Since this seems to create some confusion, you might want to add your precise question (i.e. "is $\pi_1(H)∗\pi_1(H)$ abstractly isomorphic to $\pi_1(H\cup H′)$?") to your post. –  Rasmus Feb 19 '11 at 3:31

2 Answers 2

up vote 4 down vote accepted

The two groups are not isomorphic. See Thm 1.2 of Topology Appl. 123 (2002) 479-505.

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This morphism is not surjective, for the exact same reason $\pi(H)$ is not the free group with countably many generators. Just as there are loops in $\pi_1(H)$ that go through an infinite sequence of circles, here you have loops in $\pi_1(H \cup H')$ that go through an infinite sequence of circles, and who change whose circle they're using an infinite amount of times (for example, just pick the $nth$ circle from the left if $n$ is even, and from the right if $n$ is odd). So for the same reason, the free product doesn't give you all the cycles, and that morphism is not an isomorphism.

However, I'm pretty sure you can find an isomorphism between $\pi_1(H \cup H')$ and $\pi_1(H)$, because they are homeomorphic. For example, send the circle of radius $1/n$ of the $H$ from $H \cup H'$ onto the circle of radius $1/2n$ of $H$ ; and the circle of radius $1/n$ of the $H'$ from $H \cup H'$ onto the circle of radius $1/(2n-1)$ of $H$.

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Yes, I was already aware of both of these facts. The question is whether pi_1(H) is isomorphic to its free square. –  Qiaochu Yuan Feb 19 '11 at 1:23

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