# free group represented by a 4-manifold

I want to show that any free group $G$ with finitely many ($n$) generators can be represented by a 4-manifold having fundamental group $G$.

I thought about the connected sum of n copies of $S^1 \times S^3$ since $\pi_1(S^1 \times S^3)$ is free on one generator.

Since my abstract algebra knowledge is very limited, I don't understand whether it is now trivial that $G$ is isomorphic to $n$ free products of $\pi_1(S^1 \times S^3)$

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Well, how do you compute the fundamental group of a connect sum? –  Jason DeVito Dec 20 '11 at 22:21
I think I misinterpreted your question. Is your question basically "why is the free product of $n$ $\mathbb{Z}s$ isomorphic to the free group on $n$ letters?" –  Jason DeVito Dec 20 '11 at 22:24
If so, I think it is trivial as you guessed - if $a_i$ is the generator of the $i$th $\mathbb{Z}$, consider the Free group generated by the $a_i$. Map a word in free product to the same word in the free group. –  Jason DeVito Dec 20 '11 at 22:30
@Student73: A bouquet of spheres containing more than one sphere is never a manifold because of the wedge point. –  Jason DeVito Dec 20 '11 at 22:31
But... construct a nicely embedded copy $X$ of $S^1\vee\cdots\vee S^1$ in $\mathbb R^4$ and then consider the set $M$ of points which are at distance less that $\varepsilon$ from $X$. For sufficiently small $\varepsilon$, the manifold $M$ has the hopotopy type of the original wedge of circles, so it has the correct fundamental group. Next you are going to say you wanted the manifold to be closed... –  Mariano Suárez-Alvarez Dec 21 '11 at 0:54
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