The Hawaiian earring space has notoriously complicated fundamental group, and is essentially not as simple as the wedge sum of countably many circles whose fundamental group is straightforwardly given by Van Kampen Theorem. However, I'm still struggle to understand why Van Kampen fails in this case.

I can, however, largely grasp the topological difference between the two. For instance, the wedge point of countably many unit circles is locally contractible while that of the Hawaiian earring is not. But, when checking the Van Kampen Theorem given by Allen Hatcher, I really can't find anything wrong with the earring:

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Path-connectedness is trivial. It really suffices to check the red-boxed condition. Of course, for each circle, say $C_n:=\{(x-1/n)^2+y^2=1/n^2\}$, in the earring space, we can always find an open subarc $L_n$ that deformation retracts onto $(0,0)$, the wedge point. So why isn't the theorem applicable?

  • $\begingroup$ @NajibIdrissi thanks but I know nothing about colimits. Maybe it deals with the matter at the basepoint? $\endgroup$ – Vim Jun 16 '16 at 7:35
  • $\begingroup$ Strike that, Hatcher actually proves the theorem even when the indexing set is infinite. So yes you just need van Kampen's theorem, actually. $\endgroup$ – Najib Idrissi Jun 16 '16 at 7:40

Let $H$ denote the Hawaiian earring space. Then there is a surjective continuous $\coprod_n C_n\to H$ which identifies the basepoints of all the circles together. However, this map is not a quotient map! So $H$ is not $\bigvee C_n$, since it doesn't have the quotient topology.

Here's one way to see this. For each $n$, let $U_n\subset C_n$ be a small open arc around the basepoint. Then $U_n$ is open in $C_n$ for each $n$, and $\coprod U_n$ is a saturated subset of $\coprod C_n$ with respect to the equivalence relation identifying the basepoints. So the image of $\coprod U_n$ in the quotient $\bigvee C_n$ is open. But the image of $\coprod U_n$ in $H$ is not open, since it doesn't contain a neighborhood of the basepoint (in order for it to, $U_n$ would have to be all of $C_n$ for all but finitely many $n$).

Now note that Hatcher's argument makes essential use of the assumption that $\bigvee X_\alpha$ has the quotient topology in order to assert that the set $A_\alpha$ is open in $\bigvee X_\alpha$ (this is the same argument I just gave for why the image of $\coprod U_n$ would be open in $\bigvee C_n$). So this argument does not work for $H$.

  • $\begingroup$ Thanks. I'll try to understand your "satires subset" argument. (This is admittedly the first time I've ever heard of this term, but I think I can get it right with Wikipedia). $\endgroup$ – Vim Jun 15 '16 at 17:35
  • $\begingroup$ Given a quotient map $p:X\to Y$, a subset $U\subseteq X$ is saturated if $U=p^{-1}(p(U))$. This just means it is a union of equivalence classes for the equivalence relation of the quotient. By definition of the quotient topology, if $U$ is open, this implies $p(U)$ is open in $Y$. $\endgroup$ – Eric Wofsey Jun 15 '16 at 17:47

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