Tangent space of quotient space Let $\pi : M \rightarrow M/G$ be the canonical projection, where $M$ is a manifold and $M/G$ is a quotient manifold.
Now, what can we say about $d \pi (p) : T_pM \rightarrow T_p(M/G)$? From my intuition I would say that elements in $T_p(M/G)$ consists of elements in equivalence classes defined by $T_p(G_xp)$. In case that this is true. Does anybody know how to show that $T_p(M/G)$ consists actually of these kind of elements?
 A: First, we need to be a little more careful about notation. For each $p\in M$, the map $d\pi_p$ is a surjective linear map from $T_pM$ to $T_{\pi(p)}(M/G)$ (not $T_p(M/G)$, which doesn't quite make sense). 
The tangent space $T_{\pi(p)}(M/G)$ does not "consist actually of these kinds of elements." Depending on your preferred definition, it consists either of equivalence classes of smooth curves in $M/G$, or of derivations of the space of (germs of) smooth real-valued functions on $M/G$.
However, what's interesting about this case is that there is a canonical isomorphism between the quotient vector space $T_pM/T_p(G\centerdot p)$ and $T_{\pi(p)}(M/G)$, where $G\centerdot p$ is the orbit of $p$ under $G$.  (I'm not sure what you meant by $G_xp$.) The isomorphism is induced by the surjective linear map $d\pi_p\colon T_pM \to T_{\pi(p)}(M/G)$, whose kernel (as noted by @Travis) is exactly $T_p(G\centerdot p)$. By standard linear algebra, $d\pi_p$ descends to an isomorphism from $T_pM/T_p(G\centerdot p)$ to $T_{\pi(p)}(M/G)$. Because this isomorphism is defined independently of any choices, we can canonically identify each element of $T_{\pi(p)}(M/G)$ with an element of the quotient space $T_pM/T_p(G\centerdot p)$. This is what people mean when they say that the tangent space to $M/G$ at $\pi(p)$ "is" $T_pM/T_p(G\centerdot p)$.
