Suppose we identify edges of the the unity square $[0, 1] \times [0, 1]$, as in the picture:
http://de.wikipedia.org/wiki/Datei:ProjectivePlaneAsSquare.svg
Now to compute the fundamental group, for example by van Kampen theorem I list the generators and the relations. For $\mathbb{R}P^2$ it looks like this: $<ab | abab>$.
Now, I have a difficulty understanding and arguing why $a$ itself should not be a generator. The reason is of course, that $a$ is not a loop. This is exactly the problem: if we identify edges as in the picture, all the four vertices should also be identified and then $a$ should be a loop. I think of it like this: first making a Möbius strip out of the square, we are left with only two of the former vertices, the others now being "glued" to them. Now I can't imagine how to make this strip into $\mathbb{R}P^2$ without identifying the two left-oder vertices.
Can anyone explain this?
Call the vertices $v_1,\ldots,v_4$. In the first gluing we identify $v_1$ with $v_3$ and $v_2$ with $v_4$. In the second gluing (think about it...) we have again the same identification, or alternatively, no new identifications at all.
However, I don't know if it is of any interest to you, but the fundamental group of $\mathbb{R}\mathbb{P}^2$ is much easier to compute by thinking of this space as a quotient of the $2$-sphere.