Why is the tangent space to a real projective plane two dimensional? Let P be the projective plane obtained by identifying antipode points on the unit sphere.
How to prove that the tangent space at $q \in P$ to the projective plane P is 2 dimensional?
My questions are 
1, P is not a submanifold of the Euclidean space and its tangent vectors are defined in terms of equivalence classes. How to show that there exist two linearly independent tangent vectors?
2  I hope someone can give me detailed, elementary proof without using more advanced facts--I am just starting out.
Thanks
 A: Although Nils's comment  is correct and general, I'll give you a very explicit description of the tangent space $T_q\mathbb P^2$ to the projective plane $\mathbb P^2$ at $q$.
The projective plane consists of pairs $q=\{a,-a\}$ of antipodal points $a,-a\in S^2 \; (a\cdot a=1)$,  and a tangent vector $V\in T_q\mathbb P^2$ consists of pairs of couples $V=\{(a,v),(-a,-v)\}$with $v\in \mathbb R^3$ orthogonal to $a\in S^2$ : $\: a\cdot v=0$ .    
A basis of $T_q\mathbb P^2$ consists in two linearly independant tangent vectors $V, V'\in T_q\mathbb P^2$ described as $V=\{(a,v),(-a,-v)\}, V'=\{(a,v'),(-a,-v')\}$ with $v,v'\in \mathbb R^3$ linearly independant but both orthogonal to $a$.
This makes it clear that the dimension of $T_q\mathbb P^2$ is $2$.  
Reality check
Can you compute $2V-3V'\in T_q\mathbb P^2$ ?
A: The real projective plane can be identified with tuples $(a_0,a_1,a_3) \in \Bbb{R}^3$ modulo the equivalence $(a_0,a_1,a_2) \simeq (ka_0,ka_1,ka_2)$ with $k \neq 0$ and $a_0,a_1,a_2$ not all simultaneously zero. Think about why this is the same as identifying antipodal points on $S^2$. With these, we see immediately that $\Bbb{R}P^2$ can be covered by charts $U_1,U_2,U_3$ where $U_i$ is points in $\Bbb{R}P^2$ such that the $i$-th coordinate is not zero. It is clear that $U_i$ is homeomorphic to $\Bbb{R}^2$ and so we see that each point has a neighbourhood that is homeomorphic to $\Bbb{R}^2$. So the dimension of the tangent space is $2$.
A: You have a map $p:S^2\longrightarrow \mathbb{R}P^2$ that identifies antipodal points.  Given an element $x\in S^2$, there is an open set small enough so that the restriction of $p$ to $U$ is a diffeommorphism.  This means that the induced map on tangent spaces is bijective.  And $S^2$ certainly has a two-dimensional tangent space.
