Any finitely presented group is the fundamental group of some topological space?

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I have come across a problem in topology as described in the title. Here is my intuitive construction with analogy to the fundamental group of closed surfaces. Let $G=(a_1,a_2,\dots,a_n \mid r_1,r_2,\dots,r_m)$ be a finitely presented group, then let $l^i$ be the length of each reduced word $r^i$, I define a polygon with $l^1+l^2+\dots+ l^m$ sides, and separately identify each $l^i$ sides as the reduced word $r^i$ suggests. Say, if $r^i = a_1 a_2 a_1 a_2$, then identify the consecutive $l^i$ sides by $a_1 a_2 a_1 a_2$. Then I guess that the identification space we construct is the required topological space, whose fundamental group is $G$. However, this is only an intuitive way, and I don't know whether it is right.

Moreover, can anyone explain to me the details of amalgamation when applying the Van Kampen theorem to closed surfaces? I don't understand why the amalgamation of two groups is the pushout of free products. Your help would be sincerely appreciated, and sorry for my bad typing.

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Your construction is close but not quite right. For example, suppose we start with the silly presentation $\langle x:x^2, x^2\rangle$. Then the fundamental group of your space if $\mathbb Z/4\mathbb Z$ instead of $\mathbb Z/2\mathbb Z$.

The standard construction adds one polygon per relation. Your start with a wedge of circles, one per generator, and then glue a polygon per relation. The resulting thing is sometimes called the presentation complex; googling that should find you more detailed descriptions.