Arbitrary Fundamental Group and Surfaces someone had explained to me how to construct arbitrary space $X_G$ such that $\pi_1(X_G) \cong G$, but i don't remember the end.
The idea was the following : take a presentation of the group, and then for each generator $g$ take a circle $C_g$. Then make a big wedge sum with all theses circles for create a space : $X' = \vee_{g \in G}C_g  $.
For the moment, $\pi_1(X') \cong F_G$ the free group based on $G$. 
He explain to me, that just in pasting surface between $C_g$ and $C_{g'}$ you can obtain all the relations you need, if you have taken appropriate surfaces. Then, if $X$ is the big wedge with all the necessary surfaces, we obtain $\pi_1(X_G) \cong G$.
My question is : How to choose theses surfaces ? Do you know this kind of construction ??
 A: A presentation of a group $G$ consists of generators and relations. The usual construction actually goes like this (the third step is what you're missing):


*

*Take a presentation of $G$ with generators $\{g_\alpha\}_{\alpha \in A} \subset G$ (it's not necessary to wedge a circle for every $g \in G$ and relations $\{r_\beta\}_{\beta \in B}$;

*For every generator $g_\alpha$, take a circle $C_\alpha \simeq S^1$, wedge all these to get $X'$. Then $\pi_1(X') = F_A$ is the free group on the indexing set $A$;

*For every relation $r_\beta \in F_A$, choose a loop $\gamma_\beta \in [S^1, X']$ whose homotopy class represents $r_\beta \in \pi_1(X')$. Attach a $2$-disk $D^2$ to $X'$ along $\gamma_\beta$ (ie. consider $X' \cup_{\gamma_\beta} D^2$). Then each such disk "kills" the relation $r_\beta$. Let $X$ be the space where you've attached a loop for each $\gamma_\beta$.


Then in the end, the fundamental group of $X$ will be $F_A / \langle r_\beta : \beta \in B \rangle \cong G$ because we had chosen a presentation of $G$ at the beginning.

This is corollary 1.28 in Hatcher's book Algebraic Topology, if you'd like a reference with some examples too.
