Fundamental group of a wedge sum, in general (e.g. when van Kampen does not apply)

Question

What is the fundamental group of a wedge sum in general? e.g. including the times when van Kampen cannot help us.

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The Wikipedia article on wedge sums mentions that

Van Kampen's theorem gives certain conditions (which are usually fulfilled for well-behaved spaces, such as CW complexes) under which the fundamental group of the wedge sum of two spaces $X$ and $Y$ is the free product of the fundamental groups of $X$ and $Y$.

This seems strangely worded. The conditions on the space for van Kampen are that it is

• the union of path-connected open sets $A_\alpha$,
• each contains the basepoint,
• each intersection $A_\alpha\cap A_\beta$ is path-connected,
• each intersection $A_\alpha\cap A_\beta\cap A_\gamma$ is path-connected;

then it follows that the the kernel of $\Phi$ is the normal subgroup $N$ generated by all elements of the form $i_{\alpha\beta}(\omega)i_{\beta\alpha}(\omega)^{-1}$ for $\omega\in\pi_1(A_\alpha\cap A_\beta)$, and hence $\Phi$ induces an isomorphism $\pi_1(X)\cong \ast_\alpha\pi_1(A_\alpha)/N$.

(I fix a basepoint for the rest of this question, all wedge sums are identified at this basepoint.)

Back to wedge sums. If we are looking at the wedge sum $\bigvee_\alpha A_\alpha$ where each $A_\alpha$ is path connected, then these conditions are obviously true. And since the intersection of any $A_\alpha$ is trivial, the fundamental group of the wedge sum is the free product of the groups.

The only problem that I can see arising is when $A_\alpha$ is not path connected. The fundamental group of a disconnected space $X\cup Y$ is $\pi_1(X)$ where the basepoint falls in $X$ (correct me if I am wrong, maybe I have over-simplified this). What is the fundamental group of this wedge sum, then? I would guess that it is $$\ast_\alpha\pi_1(A_\alpha^\prime),$$ where $A_\alpha^\prime$ is the component of $A_\alpha$ containing the basepoint. If so, then surely $\pi_1(A_\alpha)=\pi_1(A_\alpha^\prime)$, and there is no problem after all, the fundamental group is just the free product of each fundamental group. Is this correct (the fundamental group is always $\pi_1(\bigvee A_\alpha)=\ast\pi_1(A_\alpha)$)?

The simple example I scrawled on my piece of paper is below. I figure the fundamental group is $$\pi_1(\mathsf{Green}\vee\mathsf{Blue}\vee\mathsf{Red})=\Bbb{Z}\ast\Bbb{Z}\ast C_1=\Bbb{Z}\ast\Bbb{Z},$$ which makes sense, no loop at the basepoint can be in the leftmost or rightmost components, they must be (homotopic to a loop) in the figure eight graph.

P.S. a side question, why do we use \vee for the wedge sum rather than \wedge?

Thanks in advance :-)

Answer 1

Under certain conditions they do actually mean the existence of neighborhoods, containing the common point $x_0$ of the wedge, which do retract on $x_0$.

With this condition we are able to choose suitable open path-connected sets $A_i$ to compute the desired fundamental group.

For instance if you take the $n$ wedge of circles $S^1$ at $x_0$ you can compute its fundamental group via Van-Kampen. What is important here is that we have open neighborhoods $N_i$ containing $x_0$ for each circle $S^1$ of the wedge (just imagine a little open arc on $S^1$ which contains the center $x_0$ of the wedge) which do actually retract on $x_0$ (just shrink from both sides until you reach the center $x_0$ of the wedge).

Now you can indeed define $A_i$ to be the wedge of all $N_j$ for $j \neq i$ and wedge $S^1$ i.e. $$A_i=\left(\bigvee_{i \neq j}N_j\right) \vee S^1.$$ If you now take two different $A_i$'s, i.e. $A_i$ and $A_j$ and look at the intersection $A_i \cap A_j$ you'll see that it's simply the wedge of all $n$ neighboorhoods $N_i$ of $x_0$ and since all of them retract to $x_0$ we get $$\pi_1\left( \bigvee_{i=1}^nN_i\right) \cong \pi_1(\{x_0\})=0.$$

By a similiar argument using your retracts you can compute $$\pi_1(A_i) \cong \pi_1(S^1) \cong \mathbb Z.$$

Now by applying Van Kampen you get that the fundamental group of your wedge sum is free abelian of rank $n$.

Answer 2

Suppose $X=A_0\cup A_1$ and we have $x_0\in A_0\cap A_1$. Now the Siefert-VanKampen Theorem states $$\pi_1(X,x_0)\cong\pi_1(A_0,x_0)\ast_{\pi_1(A_0,x_0)\cap\pi_1(A_1,x_0)}\pi_1(A_1,x_0)$$ is an amalgam provided several assumptions are satisfied:

1) We must have that $A_0$ and $A_1$ are open in $X$.

2) We must have that $A_0$ and $A_1$ and $A_0\cap A_1$ are path connected.

Notice that this specializes to something like the following:

Suppose $X=A_0\cup A_1$ where $x_0\in A_0\cap A_1$ and $A_0$, $A_1$, and $A_0\cap A_1$ are open, path-connected subspaces of $X$. If $A_0\cap A_1$ is simply connected, then $\pi_1(X,x_0)\cong\pi_1(A_0,x_0)\ast\pi_1(A_1,x_0)$.

Now condition (1) explains why we need to be careful about wedge products; the subspaces $A_0$ and $A_1$ must be open in $X$. Thus your desired result about fundamental groups of wedge products has the following formulation for a sufficient condition:

Let $X=\bigvee_{\alpha}A_\alpha$ be a finite wedge product of spaces $A_\alpha$ at $x_0$. If $x_0$ has a simply connected open neighborhood $U$ such that $U\cup A_\alpha$ is open for all $\alpha$ and $A_\alpha\cap U$ is path connected, simply connected and open for all $\alpha$, then $$\pi_1(\bigvee_\alpha A_\alpha,x_0)\cong\ast_\alpha\pi_1(A_\alpha,x_0)$$

This can be proved using induction and the stated version of the SVK Theorem above.

As for wedge-versus-vee, that seems to be an unfortunate product of historical stuff...