4

The Van Kampen theorem implies that, given two path-connected (pointed) topological spaces $(X,p)$ and $(Y,q)$, we can relate the fundamental group of their wegde sum with both their fundamental groups: $$\pi_1(X\vee Y,p\vee q)\cong\pi_1(X,p)\ast\pi_1(Y,q).\qquad(\star)$$ Here, $\ast$ means the free product of groups. Note that the previous holds if we require the wedge point $p\vee q$ to have a simply-connected neighborhood in $X\vee Y$. This always happens if both $X$ and $Y$ are locally simply-connected.

Now, I am trying to construct a counter-example of the fact $(\star)$. For this, I am looking at a space that is not locally simply-connected: the Hawaiian earrings. I denote it as $H\subset{\bf R}^2$, and I take the basepoint $p\in H$ to be the “wild” point $(0,0)$. Now, I take $Y$ to be a circle of radius $2$ and centered at $(2,0)$ in ${\bf R}^2$, and I also take its basepoint $q=(0,0)$. Surely, the wedge sum $(H\vee Y,p\vee q)$ is not locally simply-connected, for the wedge point does not have a simply-connected neighborhood.

My question is: Does that provide a counter-example? My reasoning is as follows:

$\;(1)\;$ The wedge sum is homeomorphic to $H$.

$\;(2)\;$ In that case we have that $\pi_1(H\vee Y,p\vee q)\cong\pi_1(H,p)$.

$\;(3)\;$ We see that $\pi_1(H,p)\ast{\bf Z}\not\cong\pi_1(H,p)$.

I am unsure about steps $(1)$ and $(3)$ above. For $(1)$, a homeomorphism would be given by shifting every circle once to the left. For $(3)$, this is because we're adding a free generator to the (infinite) presentation of $\pi_1(H,p)$? But That does not convince me, since the free group on countably generators has the property that is is invariant under adding countably many free generators.

Please correct me wherever I am wrong!

Anthony
  • 1,379
  • 1
    Your steps (1) and (2) are correct. But (3) is wrong, or at least it is not clear why would such thing hold. For example consider the free group $F(A)$ over an infinite alphabet $A$. Then $F(A)*\mathbb{Z}$ is isomorphic to $F(A)$. But of course this doesn't answer (3) because the fundamental group of the Hawaiian earring is not free. But it is so complicated, that I doubt you'll be able to show (3), even if it is true (which I doubt). – freakish Mar 29 '22 at 14:40
  • Yes, the example you gave for $(3)$ is what I wrote about in the question, thus me suspecting it is more difficult than that. Thanks for your insight on the first two steps though! – Anthony Mar 29 '22 at 14:47
  • Do you know an explicit description of $\pi_1(H)$? Specifically, as locally eventually constant transfinite words on a countable alphabet. – Thorgott Mar 29 '22 at 15:09
  • 1
    I think asking if $\pi_1(H,h)\ast \mathbb{Z}\cong \pi_1(H,h)$ is the wrong question because you are not interested in whether there is such an isomorphism. Instead, you're asking if the specific map $\pi_1(H,h)\ast \mathbb{Z}\rightarrow \pi_1(H,h)$ given by van Kampen is an isomorphism. This is a more specific question, so may be easier to answer. Unfortunately, I don't know how to answer either question! – Jason DeVito - on hiatus Mar 30 '22 at 00:50
  • Unfortunately, no, I don't know of an explicit presentation of $\pi_1(H)$. The only thing I know is it surjects onto the Specker group. Regarding your comments: maybe there's a (way) simpler counter-example that I didn't think of? – Anthony Mar 30 '22 at 12:38

1 Answers1

2

Unfortunately I cannot answer the question whether $\pi_1(H,p)\ast{\bf Z}\not\cong\pi_1(H,p)$. The fundamental group of the Hawaiian earring is a fairly complicated object as you can see in

Morgan, John W., and Ian Morrison. "A van Kampen theorem for weak joins." Proceedings of the London Mathematical Society 3.3 (1986): 562-576.

I guess that $\pi_1(H,p)\ast{\bf Z}\cong\pi_1(H,p)$, but I do not know how to prove it.

Anyway, a counterexample to $(*)$ can be found in Fundamental group of wedge sum of two coni over Hawaiian earrings. This example is closely related to the Hawaiian earring. It shows that the wedge of two contractible spaces can have a highly non-trivial fundamental group.

Update.

Although the Hawaiian earring behaves "locally bad" at $p = (0,0)$, it is nevertheless true that $$\pi_1(H \vee Y, p \vee q) \cong \pi:1(H.p) * \pi_1(Y,q) . \tag{1}$$ The reason is that $Y$ behaves "locally nice " at $q$. If one of spaces involved in the wedge is "locally nice" at its basepoint, we can prove $(1)$.

See the answers to Under what conditions $\pi_1(X \vee Y) \approx \pi_1(X) * \pi_1(Y)$?

Paul Frost
  • 87,968