# For a free group G with finite presentation, show there exists a pointed space whose fundamental group is G

Suppose $G=\langle g_1,...,g_n|r_1,...r_m\rangle$. Show there is a pointed space $(X_G,x_0)$ with $\pi_1(X_G,x_0)=G$.

Hint: Use Van Kampen's theorem.

My attempt: First note that $G$ is the amalgamation of $G_i:=\langle g_i|r_1,...,r_m\rangle$, ie: $G= G_1\ast...\ast G_n$. Then since $G_i\cap G_j={e}$, the identity, for $i\neq j$, the pairwise intersections are trivially path connected. If we let the base point be $e$, then all I need to do to show is a that the space can be somehow represented by a union of the $G_i$, but don't know how.

• Take a peek at "bouquet" or wedge sum of circles, perhaps also under van Kampen. – DonAntonio Feb 20 '17 at 17:32
• What does $G_i$ mean if, say, $G =\langle x,y| x^2y^2, x^{-1}y x\rangle$? More specifically, if the relations $r_i$ all involve all the generators what is $G_i$? – Jason DeVito Feb 20 '17 at 17:46
• Would it then make more sense to define the $G_i=\langle g_i\rangle$ and then $G=G_1\ast...\ast G_n /\{r_1,...,r_n\}$? – George Feb 20 '17 at 22:39