Fundamental group of a triangle with the three vertices identified? Which is the fundamental group of a triangle with the three vertices identified?
I call $X$ the previous space. I take $U$ be a disk inside the triangle and $V$ the complement of a disk inside the triangle (I take them so that $U \cap V$ is homotopy equivalent to a circle). Hence $\pi_1(U \cap V)=\mathbb{Z}$, $\pi_1(U)=(0)$ and $\pi_1(V)=F^3$ is the free group on $3$ generators (i.e. it is generated by the three edges). Then, by Van Kampen 
$$
\pi_1(X)=F^3/N
$$
but I cannot express the normal subgroup $N$. My intuition says that I have to obtain $\pi_1(X)=<a,b,c: abc=1>$, i.e. the free group on two generators. Is my intuition right? Can you help me please?
 A: I'm not sure which version of van Kampen's theorem that you're using, but in case you're using the combinatorial version, then it exactly says that $\pi_1(X)=\langle x,y,z\mid i(w)=1\rangle$, where $i:\pi_1(U\cap V)\to\pi_1(X)$ is induced by the inclusion and $w\in\pi_1(U\cap V)\cong\mathbb Z$ is the generator. Here $w=[\gamma]$, where $\gamma$ is the loop around the circle in the intersection. Note that this loop is homotopic to following the three edges after each other, so that $i(w)=xyz$, which is exactly the group you mentioned.
As a side note, I envision the space $V$ as follows:

A: There is a general result involving a basic construction on groupoids: given a groupoid $G$ with object set $S$ and a function $f: S \to T$ to a set $T$ one can construct a new groupoid say $f_*(G)$ with object set $T$ and with a universal property which is nicely expressed by saying that there is the  following pushout diagram of groupoids, in which $D(S)$ is the discrete groupoid on $S$:
$$ \begin{matrix}
D(S) & \xrightarrow{D(f)} & D(T) \\
\downarrow && \downarrow\\
G & \xrightarrow{\bar{f}} & f_*(G)
\end{matrix} 
$$
This idea is due to Philip HIggins, see his downloadable book Categories and Groupoids. 
A particular case is when $T$ is a singleton, when we get the universal group  $U(G)$ of the groupoid $G$. If all the vertex groups of $G$ are trivial, then $U(G)$ is a free group. 
Special cases of this construction $U(G)$ are free products of groups, and free groups.
This result has topological applications which are given in Section 9.1 of the downloadable book Topology and Groupoids, as they were in the 1968, 1988  differently titled editions. One has to introduce the notion of the fundamental groupoid $\pi_1(X,S)$ on a set $S$ of base points to deal adequately with the situation when one is identifying  some or all of a discrete  discrete set $S$  of points. (Of course, one can get round the problem but the idea is to "match the algebra to the geometry, not force the geometry into a particular mode simply because that mode is more familiar", as remarked, I was told, by Philip Hall.) 
I published  this idea of $\pi_1(X,S)$ in 1967 in order to have a version of the van Kampen Theorem which would compute the fundamental group of the circle, which is, after all, the basic example in algebraic topology. I am unsure why this this idea is, almost 50 years later,  found  "ideosyncratic". 
Compare with this mathoverflow question and answers. 
A: I'm not sure what statement of Van Kampen you are using, but loosely speaking, $N$ is generated by loops of the form $i(\omega)j(\omega)^{-1}$ where $i$ and $j$ are induced by the inclusions $U\cap V\to V$ and $U\cap V\to U$ respectively and $\omega\in\pi_1(U\cap V)$. For a canonical generator $\omega_0\in\pi_1(U\cap V)$ (i.e. a loop that looks like a circle in $U\cap V$), we have that $i(\omega)j(\omega)^{-1}=i(\omega)=abc$ (where $a,b,c$ denote the generators for your $F^3$, which indeed correspond to the edges of the triangle; note that we can "stretch out" $\omega_0$ in $V$ to "lie on the edges"). Therefore we indeed get $\pi_1(X)=\langle a,b,c\mid abc=1\rangle$.
