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In order to show that this relation is an equivalence relation, we need to prove three properties: reflexivity, symmetry, and transitivity. Let's verify them one by one: 1. Reflexivity: The relation must hold for any \(x_1(t)\). So for \(x_1(t)=x_2(t)\), we have \(x_1(0)=x_2(0)=x_0\) and \(\mathrm{d}\left(x_1(t), x_2(t)\right)=0\), which is indeed \(o(t)\), as \(t \rightarrow 0\). Thus, the relation is reflexive. 2. Symmetry: Suppose the relation holds for \(x_1(t)\) and \(x_2(t)\). Then \(x_1(0)=x_2(0)=x_0\) and \(\mathrm{d}\left(x_1(t), x_2(t)\right)=o(t)\) as \(t \rightarrow 0\). Swapping \(x_1(t)\) and \(x_2(t)\), we also have \(x_2(0)=x_1(0)=x_0\) and \(\mathrm{d}\left(x_2(t), x_1(t)\right)=o(t)\) as \(t \rightarrow 0\). So the relation is symmetric. 3. Transitivity: Suppose that the relation holds for \(x_1(t)\) and \(x_2(t)\) and for \(x_2(t)\) and \(x_3(t)\). We have: a) \(x_1(0)=x_2(0)=x_0\) and \(\mathrm{d}\left(x_1(t), x_2(t)\right)=o(t)\) as \(t\rightarrow 0\), b) \(x_2(0)=x_3(0)=x_0\) and \(\mathrm{d}\left(x_2(t), x_3(t)\right)=o(t)\) as \(t\rightarrow 0\). We need to show that the relation holds for \(x_1(t)\) and \(x_3(t)\). We have \(x_1(0)=x_3(0)=x_0\). Furthermore, \(\mathrm{d}\left(x_1(t), x_3(t)\right) \le \mathrm{d}\left(x_1(t), x_2(t)\right) + \mathrm{d}\left(x_2(t), x_3(t)\right)\), which is \(o(t)+o(t) = o(t)\) as \(t \rightarrow 0\). Therefore, the relation is transitive. Since the relation satisfies reflexivity, symmetry, and transitivity properties, it is indeed an equivalence relation.

To establish a one-to-one correspondence between vectors \(\mathbf{v} \in T\mathbb{R}^{m}_{x_0}\) and equivalence classes of smooth paths at the point \(x_0\), we define a map \(\Phi: T\mathbb{R}^{m}_{x_0} \rightarrow\) (set of equivalence classes of smooth paths) as follows: Given a tangent vector \(\mathbf{v} \in T\mathbb{R}^{m}_{x_0}\), let \(x_1(t) = x_0 + t\mathbf{v}\). Then, define \(\Phi(\mathbf{v}) = [x_1(t)]\), the equivalence class of \(x_1(t)\). To show that this map is one-to-one (injective), we need to prove that if \(\Phi(\mathbf{v}_1) = \Phi(\mathbf{v}_2)\), then \(\mathbf{v}_1 = \mathbf{v}_2\). If \(\Phi(\mathbf{v}_1) = \Phi(\mathbf{v}_2)\), it means that \(x_1(t)\) and \(x_2(t)\) belong to the same equivalence class, so \(\mathrm{d}\left(x_1(t), x_2(t)\right) = o(t)\) as \(t \rightarrow 0\). Since \(x_1(t) = x_0 + t\mathbf{v}_1\) and \(x_2(t) = x_0 + t\mathbf{v}_2\), we can rewrite this condition as \(\mathrm{d}\left(t\mathbf{v}_1, t\mathbf{v}_2\right) = o(t)\) which implies \(\mathbf{v}_1 = \mathbf{v}_2\). Hence, the map is injective. To show that the map is onto (surjective), we need to prove that given any equivalence class of smooth paths, there exists a vector \(\mathbf{v}\) such that \(\Phi(\mathbf{v})\) belongs to that equivalence class. Let \([x_3(t)]\) be an arbitrary equivalence class of smooth paths. By the definition of equivalence classes, \(x_3(0) = x_0\) and \(\mathrm{d}\left(x_3(t), x_0\right) = o(t)\). We define \(\mathbf{v} := \dfrac{\mathrm{d}x_3}{\mathrm{d}t}(0)\). Then, \(x_1(t) = x_0 + t\mathbf{v}\) has the same tangent vector at \(t = 0\), and thus \(x_1(t)\) and \(x_3(t)\) belong to the same equivalence class. Consequently, the map is surjective. Since the map \(\Phi\) is both injective and surjective, there exists a one-to-one correspondence between vectors \(\mathbf{v} \in T\mathbb{R}^{m}_{x_0}\) and equivalence classes of smooth paths at the point \(x_0\).

We can introduce addition and scalar multiplication operations for equivalence classes of paths using the tangent space structure that we established in part (b) with the map \(\Phi\). Here's how: Addition: Given equivalence classes \([x_1(t)]\) and \([x_2(t)]\), we can find corresponding tangent vectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\) by the inverse of the map \(\Phi\). Then, we can define the sum of these equivalence classes by using the tangent space addition: \(\mathbf{v}_3 = \mathbf{v}_1 + \mathbf{v}_2\). Now, find the corresponding path of this sum of vectors by applying \(\Phi(\mathbf{v}_3) = [x_3(t)]\), where \(x_3(t) = x_0 + t\mathbf{v}_3\). Finally, define the addition of equivalence classes as \([x_1(t)] + [x_2(t)] := [x_3(t)]\). Scalar multiplication: For a scalar \(\lambda \in \mathbb{R}\) and an equivalence class \([x_1(t)]\), we can find the corresponding tangent vector \(\mathbf{v}_1\) by the inverse of the map \(\Phi\). Define the scalar multiplication of the equivalence class by using the scalar multiplication of the tangent space: \(\mathbf{v}_2 = \lambda \mathbf{v}_1\). Now, find the corresponding path of this scalar multiple vector by applying \(\Phi(\mathbf{v}_2) = [x_2(t)]\), where \(x_2(t) = x_0 + t\mathbf{v}_2\). Finally, define the scalar multiplication of equivalence classes as \(\lambda[x_1(t)] := [x_2(t)]\). These definitions allow us to use the tangent space structure when dealing with equivalence classes of smooth paths at the point \(x_0\).

The addition and scalar multiplication operations we introduced above are defined based on the tangent space structure, which is a linear space. Linear spaces are coordinate system independent, and therefore, the operations we have defined on equivalence classes of paths will also be coordinate system independent. In other words, the results obtained by using these operations will not depend on the choice of the coordinate system in \(\mathbb{R}^m\).