How to compute the fundamental group of a necklace of $\mathbb{S}^1$' s? I was trying to compute $\pi_1 (X)$ where $X =$ "necklace of $n$ $\mathbb{S}^1$'s". At first, I tried using Van Kampen theorem however I could not find open sets $U$ and $V$ such that $U \cap V$ is path connected. I tried using the covering space $E =\bigvee^\omega \mathbb{S}^1$ via the projection  as in the case of the real line covering the circumference, however I do not know how to compute $p_*( \pi_1 (E, e))$. I would appreciate an answer where one computes it using both methods (Van Kampen, maybe with an arbitrary colimit, and covering spaces).
Thanks in advance.
 A: As explained in the comments, the op means by $X$, that the right side of the first circle is glued at a point to the left side of the second circle, and the right of the second to the left of the third and so on to $n$, and then finally the right of the $n$th is glued to the left of the first.
First, stretch out the point at which the first and $n$th circles are glued together to be an interval with ends on either circle. Next, if we shrink the southern semicircles of each circle to a point, we see that this drags the ends of the stretched interval to the wedge point of $n$ circles and so adds another wedged circle. This operation is a homotopy equivalence and so $X$ is homotopy equivalent to $\bigvee_{i=1}^{n+1} S^1$ and so $$\pi_1(X)\cong\pi_1(\bigvee_{i=1}^{n+1} S^1)\cong \ast_{i=1}^{n+1} \pi_1(S^1)\cong\ast_{i=1}^{n+1}\mathbb{Z}$$ ie the $(n+1)$-fold free product of $\mathbb{Z}$ with itself.
A: The space you are talking about is a finite graph, namely a graph with $n$ vertices, and two edges joining vertices $i$ and $i+1 \pmod{n}$ for $i=1,2,\ldots,n$. To find the fundamental group of a graph, you take any spanning tree and contract it to a point (this doesn't change the homotopy tpye as the tree is contractible); what you get is a bouquet of some number $k$ of circles, and thus the fundamental group is the free group on $k$ generators. For your graph, the path of length $n-1$ that goes $1\to2\to \cdots \to n$ is a spanning tree. There are $(n-1) + 2 = n+1$ edges that are not part of the tree, so the fundamental group is free on $n+1$ letters.
