Does the pushforward of a smooth map on a manifold coincide with the derivative in a tangent direction of any local extension?

Question

Let $d\in\mathbb N$, $k\in\{1,\ldots,d\}$, $M$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary, $x\in M$, $$T_x\:M:=\{\gamma'(0):\gamma\text{ is a }C^1\text{-curve on }M\text{ through }x\}$$ denote the tangent space of $M$ at $x$, $E$ be a $\mathbb R$-Banach space and $f:M\to E$ be $C^1$-differentiable at $x$, i.e. $$\left.f\right|_{O\:\cap\:M}=\left.\tilde f\right|_{O\:\cap\:M}\tag1$$ for some $\tilde f\in C^1(O,E)$ for some $\mathbb R^d$-open neighborhood $O$ of $x$.

If I got it right, the problem with defining the derivative of $f$ at $x$ to be ${\rm D}\tilde f(x)$ is that this definition would not be well-defined, since the operator ${\rm D}\tilde f(x)$ does depend on the choice of $\tilde f$.

Now, it's easy to see that $C^1$-differentiability of $f$ at $x$ is equivalent to the $C^1$-differentiability of $f\circ\phi^{-1}$ for some $k$-dimensional chart $\phi$ of $M$ around $x$ (i.e. $\phi$ is a $C^1$-diffeomorphism from an $M$-open subset $\Omega$ onto an open subst of $\mathbb H^k:=\mathbb R^{k-1}\times[0,\infty)$).

This fact can be used to show that if $v\in T_x\:M$ and $\gamma$ is any $C^1$-curve on $M$ through $x$ with $\gamma'(0)=v$, then $${\rm D}_xf(v):=(f\circ\gamma)'(0)\tag2$$ is well-defined, i.e. independent of the choice of $\gamma$. (The crucial thing is that the derivative on a half space, like the interval on which $\gamma$ is defined or the open subset $\phi(\Omega)$ of $\mathbb H^k$ on which $\phi^{-1}$ is defined, does not depend on the choice of a differentiable local extension: Differentiability at the boundary of a function on a half space)

Now I've got two questions:

1. Can we show that, if $v\in T_x\:M$ as above, then $${\rm D}_xf(v)={\rm D}\tilde f(x)v\tag3,$$ where ${\rm D}\tilde f(x)$ is the Fréchet derivative of $\tilde f$ at $x$ and hence ${\rm D}\tilde f(x)v$ is the ordinary directional derivative of $\tilde f$ at $x$ in direction $v$?
2. If $x$ is in the topological interior $\operatorname{Int}M$ of $M$, can we conclude that $$T_x\:M=\mathbb R^d\tag4$$ and that $(3)$ holds for all $v\in\mathbb R^d$?

Regarding 2.: Since $O\cap\operatorname{Int}M$ is open, $$B_\varepsilon(x)\subseteq O\cap M\tag5$$ for some $\varepsilon>0$ and hence it should follow from $(1)$ that $f$ is actually continuously Fréchet differentiable at $x$ and $${\rm D}f(x)={\rm D}\tilde f(x)\tag6.$$ Moreover, if $v\in\mathbb R^d\setminus\{0\}$, then $$\gamma(t):=x+tv\in B_\varepsilon(x)\;\;\;\text{for }t\in I:=\left(-\frac\varepsilon{\left\|v\right\|},\frac\varepsilon{\left\|v\right\|}\right).$$ By construction, $\gamma\in C^1(I,M)$ with $\gamma(0)=x$ and $\gamma'(0)=v$ and $$(f\circ\gamma)'(0)={\rm D}f(\gamma(0))\gamma'(0)={\rm D}f(x)v\tag7$$ by the chain rule. This should yield both $(2)$ and $(3)$.

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1: If $E_i$ is a $\mathbb R$-Banach space, $\Omega_1\subseteq E_1$ and $g:\Omega_1\to E_2$, then $f$ is called $C^1$-differentiable at $x_1\in\Omega_1$ if $$\left.g\right|_{O_1\:\cap\:\Omega_1}=\left.\tilde g\right|_{O_1\:\cap\:\Omega_1}$$ for some $\tilde g\in C^1(O_1,E_2)$ for some $E_1$-open neighborhood $O_1$ of $x_1$. $f$ is called $C^1$-differentiable if it is $C^1$-differentiable at $x_1$ for all $x_1\in\Omega_1$, which is equivalent to $$g=\left.h\right|_{O_1}$$ for some $h\in C^1(O_1,E_2)$ for some $E_1$-open neighborhood $O_1$ of $\Omega_1$.

2: $\gamma$ is called $C^1$-curve on $M$ through $x$ if $\gamma:I\to M$, where $I$ is a nontrivial interval with $0\in I$, is $C^1$-differentiable with $\gamma(0)=x$.

Answer 1

Partial answer to 1.:

As in $(1)$, since $f$ is $C^1$-differentiable at $x$, $$\left.f\right|_{O\:\cap\:M}=\left.\tilde f_1\right|_{O_1\:\cap\:M}\tag8$$ for some $\tilde f_1\in C^1(O_1,E)$ for some $\mathbb R^d$-open neighborhood $O_1$ of $x$. Moreover, since $\Omega$ is $M$-open, $$\Omega=\Omega_2\cap M\tag9$$ for some open subset $O_2$ of $\mathbb R^d$. Note that $$O:=O_1\cap O_2$$ is an $\mathbb R^d$-open neighborhood of $x$ and $$\tilde f:=\left.\tilde f_1\right|_O\in C^1(O,E)\tag{10}$$ with $$\left.f\right|_{\tilde\Omega}=\left.\tilde f_1\right|_{\tilde\Omega},\tag{11}$$ where $$\tilde\Omega:=O\cap M\subseteq\Omega\tag{12}.$$

Now let $v\in T_x\:M$ and $\gamma:I\to M$ be a $C^1$-curve on $M$ through $x$ with $\gamma'(0)=v$. Since $\tilde\Omega$ is $M$-open and $\gamma$ is continuous, $$A:=\gamma^{-1}(\tilde\Omega)$$ is $I$-open and hence $$A=B\cap I\tag{12}$$ for some open subset $B$ of $\mathbb R$. So, $$(-\varepsilon,\varepsilon)\subseteq B\tag{13}$$ for some $\varepsilon>0$- Clearly, $$\tilde I:=(-\varepsilon,\varepsilon)\cap I$$ is a nontrivial intervial with $0\in\tilde I$ and $$\gamma(\tilde I)\subseteq\tilde\Omega\tag{14}.$$ Let $\tilde\gamma:=\left.\gamma\right|_{\tilde I}$. Then, $$f\circ\tilde\gamma=\tilde f\circ\gamma\tag{15}$$ and hence $$(f\circ\tilde\gamma)'(0)=(\tilde f\circ\tilde\gamma)'(0)={\rm D}\tilde f(x)\tilde\gamma'(0)={\rm D}\tilde f(x)v\tag{16}.$$

Thus, ${\rm D}_xf(v)$, as defined in the question, really coincides with ${\rm D}\tilde f(x)v$. It would be great if someone could check whether I made some mistake.