Van Kampen on an infinite wedge of circle. Let $X=S^1\lor S^1\lor\cdots$ be the countably infinite wedge of $S^1$'s. We know in the finite case, given $k$ factors $S^1$ in the wedge, that $\pi_1(X)=\Bbb Z*\Bbb Z*\cdots$ with the product taken $k$ times, and you can easily show it using Van Kampen. My question is how to deal with the infinite case. My claim is that $\pi_1(X)=\Bbb Z*\Bbb Z*\cdots$, the free product of countably infinite $\Bbb Z$. By now I have rigorously shown that $\pi_1(X)$ is not trivial using van Kampen and a similar strategy as in the finite case. But can one prove the result properly?
 A: First, Hatcher proves Van Kampen for arbitrarily many open sets, and I think it should be proven this way, as it doesn't take much more effort (see Theorem 1.20 in Hatcher's book or if you feel like (presumably) putting in a bit more work, the chapter on Van Kampen in May's book).
In any case, if you prefer sticking to Van Kampen for finite open sets, there is an easy argument to show that $\pi_1(X)\cong*_{n\in\mathbb{N}}\mathbb{Z}$, using a trick which is actually quite often useful. First, you said you managed to prove that $\pi_1(X)$ is non trivial, so I assume you will agree that the group morphism $*_{n\in\mathbb{N}}\mathbb{Z}\hookrightarrow \pi_1(X)$ defined by sending each generator of the free group into a loop around a circle in the wedge product (of course, you assign exactly one generator to each circle) is injective. Then we only need to prove surjectivity of our morphism, which we can do in the following way. An element $\alpha$ in $\pi_1(X)$ is represented by a map $[0,1]\to X$, and since $[0,1]$ is compact such a map must have compact image, thus the image must be contained in a finite number of circles in $X$. Van Kampen in the finite case shows that there is some element in $*_{n\in\mathbb{N}}\mathbb{Z}$ which is mapped to $\alpha$. This shows that $\pi_1(X)\cong*_{n\in\mathbb{N}}\mathbb{Z}$.
