Prove that $C:\mathbb{R}_x^n \to T_xU$ defined as $v \to D_v$ is a bijection 
Prove that the map $C:\mathbb{R}_x^n \to T_xU$ defined as $C(v)=D_v$ is bijection. Where
$D_v(f)=\frac{d}{dt} |_{t=0} f(x+tv)=\sum_{i=1}^{n} \frac{\partial}{\partial x} (x)v^i$
for $f \in C^\infty (U)$.



*

*Show that $C$ is a linear map.


I know how to prove that.



*Show that $C$ is injective.


If for all $f\in C^\infty (U)$, $D_v(f)=0$ that means
$\frac{d}{dt} |_{t=0} f(x+tv)=\sum_{i=1}^{n} \frac{\partial}{\partial x} (x)v^i=0$
Since this is a derivative that equal zero, does that mean $f(x+tv)$ is constant?



*Show that $C$ is onto. Hint: use Taylor's Theorem.


Not sure either.
Any help would be appreciated. Thanks
 A: First off, $f(x+tv)$ having derivative 0 at $t=0$ does not mean that $f$ is constant. For instance, take $f(t)=t^2$, then $f'(0)=0$ but $f$ is not constant. You need the derivative to vanish on the whole domain to conclude that $f$ is constant.
Anyways, let's turn to injectivity of this map. Suppose $D_v = 0$. Then for every $f \in C^\infty (U)$, $D_v(f) = 0$. So let's define $f(w) = \langle w - x, v\rangle$. The brackets here denote the usual inner product, and I'll leave checking smoothness up to you. Essentially, $f$ comes from projecting $w$ to the $v$ direction. We compute $f(x + tv) = \langle tv, v\rangle = t ||v||^2$. Hence, the derivative of this function at $t=0$ is $||v||^2$. As we assumed that $D_v = 0$, we have that $||v||^2 = D_v(f) = 0$. Hence, $v = 0$ so the kernel of this map is trivial.
For surjectivity, could you specify what exactly your definition of $T_x U$ is? The hint is correct that you have to apply Taylor's theorem, but the specifics of how this argument will be written will depend on this definition.
EDIT: Let's let $Y \in T_x U$. Let's also define by $p_i \in C^\infty (U)$, by $p_i(w) = \pi_i(w - x)$ where $\pi_i$ is projection onto the $i^{th}$ coordinate. We can therefore consider the values $\lambda_i = Y(p_i)$. Let $X = \sum \lambda_i D_{e_i}$, which is in the image of $C$. We claim $X = Y$. Indeed, take some $f \in C^\infty (U)$. By Taylor's theorem, we will write $f(w) = f(x) + \sum a_i p_i(w) + \sum (a_{ij} p_i p_j)(w)$. Here, the $a_i$ are constants and the $a_{ij}$ are smooth functions. Let $Z = X - Y$. Then $Z(f) = Z(f(x)) + \sum a_i Z(p_i) + \sum Z(a_{ij} p_i p_j)$. By the Leibniz rule, $Z(f(x)) = 0$. Furthermore, $Z(p_i) = X(p_i) - Y(p_i) = \lambda_i - Y(p_i) = 0$. Now, we are left to consider $Z(a_{ij} p_i p_j)$. Again, applying the Leibniz rule, this is $Z(a_{ij} p_i) p_j(x) + (a_{ij}p_i)(x) Z(p_j) = 0$. Hence, we have shown that $Z = 0$ so $X = Y$. We chose $X$ to be in the image of $C$, and $Y \in T_x U$ was arbitrary. Hence, $C$ is onto.
