Tangent Space is Vector Space

Prove that the operation of linear combination, as in Definition 2.2.7, makes T_p(U) into an n-dimensional vector space over \mathbb{R}. The zero vector is the infinitesimal curve represented by the constant p. If \langle s\rangle_p\in T_p(U), then -\langle s\rangle_p=\langle s^-\rangle_p where s^-(t)=s(-t), defined for all sufficiently small values of t.

Proof:

We verify the axioms of a vector space.

Multiplicative axioms:
1\langle s_1\rangle_p=\langle 1s_1+0-(1+0-1)p\rangle_p=\langle s_1\rangle_p
(ab)\langle s_1\rangle_p=\langle abs_1-(ab-1)p\rangle_p
\begin{aligned}  a(b\langle s_1\rangle_p)&=a\langle bs_1-(b-1)p\rangle_p\\  &=\langle abs_1-(ab-a)p-(a-1)p\rangle_p\\  &=\langle abs_1-(ab-1)p\rangle_p\\  &=(ab)\langle s_1\rangle_p  \end{aligned}

Additive Axioms:
\langle s_1\rangle_p+\langle s_2\rangle_p=\langle s_2\rangle_p+\langle s_1\rangle_p=\langle s_1+s_2-p\rangle_p
\begin{aligned}  (\langle s_1\rangle_p+\langle s_2\rangle_p)+\langle s_3\rangle_p&=\langle s_1+s_2-p\rangle_p+\langle s_3\rangle_p\\  &=\langle s_1+s_2-p+s_3-p\rangle_p\\  &=\langle s_1+s_2+s_3-2p\rangle_p  \end{aligned}
\begin{aligned}  \langle s_1\rangle_p+(\langle s_2\rangle_p+\langle s_3\rangle_p)&=\langle s_1\rangle_p+\langle s_2+s_3-p\rangle_p\\  &=\langle s_1+s_2+s_3-2p\rangle_p  \end{aligned}
\langle s\rangle_p+\langle s^-\rangle_p=\langle s+s^- -p\rangle_p

\frac{d}{dt}f(s(t)+s(-t)-p)|_{t=0}=0=\frac{d}{dt}f(p)|_{t=0}

Hence \langle s\rangle_p+\langle s^-\rangle_p=\langle p\rangle_p.
\langle s_1\rangle_p+\langle p\rangle_p=\langle s_1+p-p\rangle_p=\langle s_1\rangle_p

Distributive Axioms:
\begin{aligned}  a(\langle s_1\rangle_p+\langle s_2\rangle_p)&=a\langle s_1+s_2-p\rangle_p\\  &=\langle a(s_1+s_2-p)-(a-1)p\rangle_p  \end{aligned}
\begin{aligned}  a\langle s_1\rangle_p+a\langle s_2\rangle_p&=\langle as_1-(a-1)p\rangle_p+\langle as_2-(a-1)p\rangle_p\\  &=\langle a(s_1+s_2)-2(a-1)p-p\rangle_p\\  &=\langle a(s_1+s_2-p)-(a-1)p\rangle_p\\  &=a(\langle s_1\rangle_p+\langle s_2\rangle_p)  \end{aligned}
(a+b)\langle s_1\rangle_p=\langle (a+b)s_1-(a+b-1)p\rangle_p

a\langle s_1\rangle_p+b\langle s_1\rangle_p=\langle as_1+bs_1-(a+b-1)p\rangle_p=(a+b)\langle s_1\rangle_p

Hence T_p(U) is a vector space over \mathbb{R}. Since U\subseteq\mathbb{R}^n, T_p(U) is n-dimensional.

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