Tangent vector and the cotangent space

Question

I tried to prove that $\{dx^i\}$ is the basis of the cotangent space $T_pM$ of a manifold $M$ for $x^1,\dots,x^n$ local coordinates in neighborhood $U$ of $p\in M$.

I reed somewhere that $dx^iX_p=X_p(x^i)$. I don't see why this holds.

We defined tangent vectors $X_p$ as linear derivations at $p\in M$. That means that $X_p:C^\infty(M)\to\mathbb{R}$ satisfying

1. $X_p(af+bg)=aX_p(f)+bX_p(g)$
2. $X_p(fg)=g(p)X_p(f)+f(p)X_p(g)$

$\forall a,b\in\mathbb{C}$ and $\forall f,g\in C^\infty(M)$.

Thanks for your help

Answer 1

I reed somewhere that $dx^iX_p=X_p(x^i)$. I don't see why this holds.

$x^i:M\to \mathbb R$ are coordinate functions and the $dx^i$ are their derivative maps (sometimes called "differentials") which take a tangent vector at a point $p$ as argument. A formula is given to calculate it as: $$dx^i(X_p)=X_p(x^i),$$ where I took the liberty of using brackets in the LHS instead of your riskier juxtaposition. The RHS is known to you, and is also what's called a directional derivative of the function $x^i.$