The fundamental group of a pair of Hawaiian earrings 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 $H$.
 A: The two groups are not isomorphic. 
See Thm 1.2 of Topology Appl. 123 (2002) 479-505.
A: 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 $H$.
