# Fundamental group of the topological space obtained by identifying the four vertices of a square

The task is: Compute the fundamental group of the topological space obtained by identifying the four vertices of a square.

So we identify the vertices with the same letter. Can we say something about the orientation of sides ?

How can we use Van-Kampen theorem here ?

• This questions is very unclear. Are you literally asking for the space where all vertices are identified but nothing is done to the edges? – Cheerful Parsnip Nov 20 '13 at 14:03
• Yes, this is exactly the exercise. – derivative Nov 20 '13 at 14:04
• Please ensure that your question is fully understandable without the title. The title's purpose is to attract people to your question (and to make it easily searchable). Once it has done that, it's job is done. The question should be self-contained. – Lord_Farin Nov 20 '13 at 15:48

Or to use Van Kampen, a neighborhood of the perimeter has free fundamental group generated by the four edges. Gluing in the interior of the square adds the relation that the product of these four generators is trivial. So we have $\langle a,b,c,d\,|\, abcd=1\rangle$, which isomorphic to a free group on three generators.
I like to see this as part of a general question: Let $X$ be a topological space with a subspace $A$ which is discrete and $i:A \to X$ is a cofibration. Let $f:A \to B$ be a function to another set, and let $Y$ be the space obtained from $X$ by identifying the points of $A$ according to $f$. (For example, $B$ could be a singleton.) We wish to describe the morphism $\phi: \pi_1(X,A) \to \pi_1(Y,B)$ of groupoids. It is a consequence of the groupoid version of the Seifert-van Kampen Theorem that $\phi$ is a universal morphism in the sense of Higgins' book Categories and Groupoids. If $B$ is a singleton, then we get the "universal group" of the groupoid. The universal group of a groupoid gives the left adjoint to the inclusion of the category of groups into the category of groupoids.
In particular, if $X$ is contractible, then the universal group of $\pi_1(X,A)$ is a free group.