Is my proof that we can endow the tangent space with a vector space structure alright? First some initial definitions (because I am working from a physics point of view rather than from a mathematician's point of view, and I suspect the way this problem is approached will vary slightly between the two disciplines):


*

*$\gamma:\mathbb{R}\supseteq U\to M$ is a curve on $M$ if
$\gamma$ is differentiable.


*The velocity, $v_{\gamma,p}:C^{\infty}(M)\to\mathbb{R}$ is a
linear function defined as
$v_{\gamma,p}=d_{\lambda}(f\circ\gamma)|_{\lambda_0}$, where
$\gamma$ is a function in $\lambda$ and $p=\gamma(\lambda_0)$. Note
$d_\lambda=\frac{d}{d\lambda}$.

I am trying to prove that the tangent space at a point $p$ on a manifold $M$ defined by $$T_p(M)=\{v_{\gamma,p}:\gamma\text{ is a curve passing through $p$ parametrized by }\lambda\},$$ can be endowed with a vector space structure by the defining the following as the addition and scalar multiplication operations on the space:
$$(v_{\gamma,p}\oplus v_{\delta,p})(f)=v_{\gamma,p}(f)+v_{\delta,p}(f)$$
$$\alpha\cdot v_{\gamma,p}(f)=\alpha v_{\gamma,p}(f)$$
The problem is that we need to verify these operations are closed on $T_p(M)$. I have been following this video by Frederic Schuller (timestamp included!), and I find my approach to be different than his. His proof for closure under scalar multiplication is as follows:

We will prove there exists a curve $\tau$ so that $\alpha\cdot v_{\gamma,p}(f)=v_{\tau,p}$. First, suppose $\gamma: U\to M$. We need to ensure that there exists a $\lambda$ in the domain of $\tau$ so $\tau(\lambda)=\gamma(\alpha\lambda)$. We note that $\gamma(\lambda_0)=p$, so we choose $\tau$ such that $\tau=\gamma(\alpha\lambda+\lambda_0)$, such that $\tau(0)=\gamma(\lambda_0)=p$, such that $\gamma$ passes through $p$. For simplicity, we may as well assume that $U$ is an open interval in $\mathbb{R}$ containing $\lambda_0$. i.e. $U=(a,b)\ni \lambda_0$. We may well, set the domain of $\tau$ to then be $(a-\lambda_0,b-\lambda_0)\ni 0$. Moreover, we will define an auxiliary function $\mu_{\alpha}:\mathbb{R}\to\mathbb{R}$ defined by $\lambda\overset{\mu_{\alpha}}{\mapsto}\alpha\lambda+\lambda_0$. In particular then $\tau(\lambda)=\gamma\circ\mu(\lambda)$. Now, we verify that the required property holds: We have $$d_{\lambda}(f\circ\tau)|_{0}=d_{\lambda}((f\circ\gamma)\circ\mu)|_0.$$ Applying chain rule, we have that this is $$(f\circ\gamma)'|_{\mu(0)=\lambda_0}\cdot\mu'(0)=\alpha(f\circ\gamma)'|_{\lambda_0}=\alpha v_{\gamma,p},$$ as required.

My approach is slightly different, and I am curious to know if it works as well. I think the problem with my formulation may lie in ensuring that $p$ lies in the curve $\tau$. Nevertheless, here it is:

Let $\tau=\gamma(\alpha\lambda)$. To ensure $p$ lies in the image of $\tau$, we set the domain of $\tau$ to include $\frac{\lambda_0}{\alpha}$. In particular, if $U=(a,b)\ni\lambda_0$ is the domain of $\gamma$, set the domain of $\gamma$ to be $V=(c,d)$, where $\frac{\lambda_0}{\alpha}\in V$ If we do this, we have to treat the case where $\alpha=0$ separately. Again, we can define an auxiliarly map $\mu_{\alpha}$ defined by $\mu_{\alpha}(\lambda)=\alpha\lambda$. Now we have $$d_{\lambda}(f\circ\tau)|_{\frac{\lambda_0}{\alpha}}=d_{\lambda}(f\circ\gamma\circ\mu)|_{\frac{\lambda_0}{\alpha}}.$$ We can then follow the same procedure to obtain an identical result to the above. When $\alpha=0$, I'm not sure how to proceed. Previously, I considered setting $\tau$ to be a "zero" map, but there's no real sense of $0\in M$.

To me, my approach seems slightly less technical and easier to write down (though there is that added case that I haven't been able to finish the proof for) than the approach taken by Schuller, which makes me wonder if there's anything wrong with it. Can we finish my proof for the case where $\alpha$ is $0$? Or is there something fundamentally wrong with my approach? Thoughts?
Thanks in advance for your help -- it is very much appreciated!
Edit: I suspect if we let $\tau(\lambda)=\gamma(\mu(\lambda))$, where $\mu(\lambda)=\lambda_0$ for any $\lambda$, then the case for the proof where $\alpha=0$ will work. Indeed, we have $$d_{\lambda}(f\circ\tau)|_0=d_{\lambda}(f\circ\gamma\circ\mu)_0=(f\circ\gamma)'|_{\mu{0}=\lambda_0}\cdot\mu'|_0=0=0\cdot v_{\tau,p}(f),$$ as required. Now this seems alright to me, and indeed $\tau(0)=p$, so $\tau$ is a "curve" passing through $p$. This seems okay to me aside from the fact that this $\tau$ isn't really a curve. Physically though, this makes sense, if we interpret the curve as a "direction" and velocity as a directional derivative, as we often do in physics.
 A: Your definition works. If $r \ne 0$, then multiplication by $r$ gives us a diffeomorphism $\mu_r : \mathbb R \to \mathbb R, \mu_r(\lambda) = r \lambda$. Thus for $\alpha \ne 0$ the set $U_\alpha = \mu_{1/\alpha}(U)$ is an open subset of $\mathbb R$ and
$$\tau : U_\alpha \to M, \tau(\lambda)  = \alpha \lambda$$
is a well-defined curve. You have $\tau(\lambda_0/\alpha) = \gamma(\lambda_0) = p$ and you correctly show that $v_{\tau,p} = \alpha v_{\gamma,p}$.
However, I think that the definition of $v_{\gamma,p}$ in your question is dissatisfactory. In fact, the point $\lambda_0$ is an essential ingredient here. It is not sufficient to consider curves $\gamma : U \to M$ with $p \in \gamma(U)$, we must consider "basepoint-preserving curves" $\gamma : (U,\lambda_0) \to (M,p)$. These are curves $\gamma : U \to M$ mapping a given point $\lambda_0 \in U$ to $p \in M$. In your definition you say

The velocity, $v_{\gamma,p}:C^{\infty}(M)\to\mathbb{R}$ is a linear function defined as $v_{\gamma,p}=d_{\lambda}(f\circ\gamma)|_{\lambda_0}$, where $\gamma$ is a function in $\lambda$ and $p=\gamma(\lambda_0)$.

Although it makes clear that we require $p=\gamma(\lambda_0)$ for some $\lambda_0 \in U$, it does not make clear which $\lambda_0$ we shall take. If $\gamma$ is not injective, then we may have many such points $\lambda_0$ and the values $d_{\lambda}(f\circ\gamma)|_{\lambda_0}$ may differ. Thus let me emphasize again that $\lambda_0$ belongs to the specification of a curve through $p$ if we want to define $v_{\gamma,p}=d_{\lambda}(f\circ\gamma)|_{\lambda_0}$.
Clearly any differentiable map $\phi : (\tilde U,\tilde\lambda_0) \to (U,\lambda_0)$ with derivation $\phi'(\tilde \lambda_0) = 1$ gives us a curve $\tilde \gamma = \gamma \phi:  (\tilde U,\tilde\lambda_0) \to (M,p)$ such that $v_{\tilde \gamma,p} = v_{\gamma,p}$. You can in particular take any translation $\theta$ on $\mathbb R$ ($\theta(t)  = t - t_0$) to shift $(U,\lambda_0)$ arbitrarily: Consider $\phi : (\theta^{-1}(U), \lambda_0 +t_0) \stackrel{\theta}{\to} (U,\lambda_0)$. This shows that one can restrict to curves $\gamma : (U,0) \to (M,p)$ to define $v_{\gamma,p}$. This is a standard approach, but it is of course a matter of taste.
For $\alpha = 0$ the map $\alpha v_{\gamma,p}$ is the zero map which has the form $v_{c,p}$ with the constant map $c : (\mathbb R,0) \to (M,p)$.
