As I read an algebraic topology book, I felt I knew exactly what the fundamental group is geometrically! I thought it counts the number of independent cycles. (my definition of dependence cycles (that may be incorrect): $\alpha,\beta$ are two dependent cycles if $\exists m,n\in\Bbb Z$ s.t. $\alpha^n=\beta^m$.)
And according to this explanation it was always a serious question for me that why almost all authors compute the fundamental group of circle in very long way, some of them in a separate chapter and after proving many theorems instead of observing that there is only one independent cycle in circle! After these doubts, I read that $\pi_1(\Bbb RP^2)=\Bbb Z_2$. And here was the end of my dream. Since in term of my interpretation, the fundamental group is always free product of some $\Bbb Z$ but $\pi_1(\Bbb RP^2)=\Bbb Z_2$ is not in that form! So I want to know that
What does fundamental (homotopy) groups measure?
Isn't it true that if the space does not have a trivial loop, then it must contain at least one generator?
I think my interpretation is partially true but it ignores some information of the space. Or perhaps the missed point of my interpretation is that I don't consider the base point to be fixed.
The above interpretation also can be applied for $\pi_2$ (and $\pi_n$). i.e. $\pi_2$ counts the number of independent topological spheres (But in this case I don't know what is the independent topological spheres!!).
Maybe instead of general ideas, look at simple examples where the fundamental group is not a free product. Let's look at a torus, represented as a square with opposite sides identified. Each of the sides $a$ and $b$ represent incontractible loops, but if you go around the square, then the loop $aba^{-1} b^{-1}$ can be contracted (shrinking the boundary of the square into a mid-point).
Following picture illustrates a homotopy between $ab$ and $ba$ in a torus.
With $\Bbb{RP}^2$, represent it as a ball with lower half-circle $a$ identified with upper half-circle $a$. Then there is no way to contract $a$ but of course, $aa$ goes around the circle and can be contracted.
Nullhomotopy of $aa$:
Let's try to nullhomotop just $a$.. intuitively, you can probably see you don't succeed
To illustrate that fundamental group of $\mathbf{SO}(3)$ is $\Bbb Z_2$ is even more intuitive. Take a glass of beer, hold it in your right arm. Rotate it by 360 degrees, with your shoulder being "fixed" (and don't pour anything out). Your hand is "twisted" and represent a loop in $\mathbf{SO}(3)$. Move your hand up and continue twisting in the same rotation; after another 360 degrees, your hand is untwisted again. Great video.
If you want to think in terms of "independence" -- think of generators and relations and note that $\Bbb Z=\langle \alpha| \varnothing\rangle$ (empty relation) and $\Bbb Z_2=\langle \alpha| \alpha^2=1\rangle$. For $\pi_1$, you can restrict yourself to the 2-skeleton. Homotopy group of $1$-skeleton is indeed a free group of generating cycles. Every two-cell adds one relation.