If $(f_*)_p(v_p) = (f_*)_p(w_p)$ for all $f\in C^\infty(M)$, then $v_p = w_p$.

Question

Let $M$ be a smooth manifold. Let $p \in M$ be fixed and $v_p, w_p \in T_p M$. Suppose that

$$(f_*)_p(v_p) = (f_*)(w_p)$$

for all smooth maps $f: M \to \mathbb{R}$ (here $(f_*)_p: T_pM \to T_{f(p)}\mathbb{R}\cong \mathbb{R} $ denote the differential of $f$).

Can we conclude that $v_p = w_p$.

Obviously, the answer is yes. I simply have to pick the right smooth map $f$. I think this will somehow include bump functions or something like that.

Answer 1

By working in a chart, it is no restriction to assume that $M$ is an open subset of $\mathbb R^n$, and then the tangent space at $p$ is naturally identified with $\mathbb R^n$. The differential at $p$ of (the restriction to $M$ of) a linear map $f : \mathbb R^n \to \mathbb R$ is just $f$ itself.

So we have that $v_p, w_p \in \mathbb R^n$ are such that $f(v_p) = f(w_p)$ for every linear map $f : \mathbb R^n \to \mathbb R$. Hence $v_p = w_p$.

Answer 2

Explicitly in coordinates $(x^i)$, write $v= v^i\frac{\partial}{\partial x^i}, w = w^i\frac{\partial}{\partial x^i}$, and let $f=x^j:U \to \mathbb R$ be the coordinate function for some fixed $j$ (You can use a bump function to extend the domain to all of $M$). Then the differential of $f$ is exactly $$f_* = df = dx^j$$ (the same as the dual 1-form to $\frac{\partial}{\partial x^j}$).

So (at $p$) $$v^j = dx^j(v) = f_*(v) = f_*(w) = dx^j(w) = w^j$$ for every $j$.