Fundamental group of a bouquet of circles

Question

Consider a bouquet of $n$ circles, centered at $(1,0), (2,0),...,(n,0)$ $$ W_n= \bigcup_1^nC_r$$ with $$C_r:(x-r)^2+y^2=r^2$$

I want to compute the fundamental group of this space.

$W_n$ is path connected. Moreover I know $\pi_1(W_1)=\pi_1(C_1)=\mathbb{Z}.$ Seeing $W_n= W_{n-1}\cup C_n$ we have both $W_{n-1}$ and $C_n$ are path connected and open. Moreover $W_{n-1} \cap C_n $ is just the origin which has trivial fundamental group. Hence $$\pi(W_n)=\pi(W_{n-1})*\pi(C_1)=\pi(W_{n-1})*\mathbb{Z}=...=\pi_1(W_1)*\mathbb{Z}...*\mathbb{Z}=$$$$\pi(C_1)*\mathbb{Z}*...*\mathbb{Z}=\mathbb{Z}*...*\mathbb{Z}$$ so it is the $n$-times free product of $\mathbb{Z}$ with itself.

Does it work? First I want to know if my solution is correct and only secondly a different one thanks.

Answer 1

Your proof is correct but a bit sloppy. It is true that the sets $W_{n-1}$ and $C_1$ are path-connected, but they are not open. Clearly no real harm has occured, but you should be careful to meet the requirements of the exact statement of the Seifert-van Kampen Theorem which you clearly applying.

The space $W_n$ is a wedge of $n$ circle $\vee^nS^1$. There are slight more sophisticated van Kampen Theorems which can compute the fundamental group of such spaces directly (i.e. by working with a family of many path-connected open sets and their intersections rather than just two).

Another way to compute its fundamental group is to construct its universal cover. There is a standard way to describe the universal covers of wedge sums. The case for $n=1,2$ is clear, but I don't even want to think about what the space looks like for higher $n$.

Answer 2

Your answer is wrong, for example, it is not very difficult to see that this fundamental group is not abelian. A way to compute that is to show that the space $W_n$ is homotopy equivalent to the wedge sum of $n$ circles and then use the Seifert-Van Kampen theorem to show that its fundamental group is the free group with $n$ generators.