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

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 :-)


2 Answers 2


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 $n$.

  • 1
    $\begingroup$ Thanks for the answer. Is there any example where we cannot find the described neighborhoods? Perhaps then the fundamental group of the wedge sum would be different to the free product of the fundamental groups of the subspaces? $\endgroup$
    – Szmagpie
    Apr 26, 2016 at 23:36
  • $\begingroup$ Yes, I assume there is, but I don't have one at disposal. I'll look for one later. $\endgroup$
    – noctusraid
    Apr 27, 2016 at 7:03
  • 2
    $\begingroup$ Having thought about it, I think the Hawaiian earring could cause problems since every open neighborhood would contain an embedded circle of radius $1/n$ for sufficiently large $n$. Let me know if that's wrong $\endgroup$
    – Szmagpie
    Apr 27, 2016 at 11:42

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...

  • $\begingroup$ He doesn't need induction as he has already the theorem in the more general form. $\endgroup$
    – noctusraid
    Apr 26, 2016 at 22:51

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