This is discussed in the standard textbooks on algebraic topology. we just pick Pick a presentation of the group G= $G = \langle g_1,g_2,...,g_n|r_1,r_2,...r_m \rangle$ where g_i $g_i$ are generators and r_j $r_j$ are relations. then Then we have a wedge of n $n$ circles and attach two cells two-cells to the wedge sum according to the relations r_j.denote $r_j$. Denote the final space X. then Van-Kampen told us Pi_1(X)=G. while $X$. Then van Kampen says $\pi_1(X)=G$. While usually X $X$ is not a manifold. , it is well-known that every finite finitely generated group G $G$ can be realized as the fundamental group of some 4-manifold X,can some one $X$. Can someone sketch the proof?besides,if X proof? Also, if $X$ could not be some manifold of dim<4,what dimension $<4$, what is the obstruction?