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The problem is about (topological) orientation of $\mathbb{R}^n$:

Define an equivalence relation on orthonormal frames in $\mathbb{R}^n$ by declaring two frames equivalent if the matrix expressing one in terms of the other has determinant $+1$. Set up an explicit correspondence between generators of $H_n (\mathbb{R}^n,\mathbb{R}^n-\{0\})$ and equivalence classes of frames.

How can I get this? Actually I need to know what kind of correspondence I need to find.

I can use the fact that the homomorphisms $$ H_q^\sharp (S^n) \rightarrow H_q^\sharp(S^n,E_n^-) \leftarrow H_q^\sharp(E^n,S^{n-1})\rightarrow H_{q-1}^\sharp(S^{n-1})$$ are (natural) isomorphisms for all $q \ge0$, $n \ge 1$. Here $H_q^\sharp$ denotes the $q$-th reduced singular homology.

Thank you for your help!

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Every ordered basis (no orthogonality or length restrictions needed) $b=(b_1,\ldots,b_n)$ defines a non-degenerate $n$-simplex $\sigma_b$ with $0$ in its interior by defining the last vertex to be $b_0 := -\sum_i b_i$. Since $0$ is in the interior, the boundary is completely in $\mathbb{R}^n\setminus\{0\}$. Therefore $\sigma_b$ induces a homology class $[\sigma_b]\in H_n(\mathbb{R}^n,\mathbb{R}^n\setminus\{0\})$. In fact it is a generator of that group because it is the image of $[id]\in H_n(\Delta_n,\partial\Delta_n)$ under the isomorphism $$H_n(\Delta_n,\partial\Delta_n) \to H_n(\Delta_n,\Delta_n\setminus\{p\}) \to H_n(\sigma_b,\sigma_b\setminus\{0\}) \to H_n(\mathbb{R}^n,\mathbb{R}^n\setminus\{0\})$$ The left arrow is an isomorphism by homotopy, the middle one is induced by an homeomorphism, the right one is an isomorphism by excision.

Now one can show that $[\sigma_{Ab}] = sgn(det(A))[\sigma_b]$ for all $A\in GL_n(\mathbb{R})$ and therefore this class is constant on the equivalence classes of frames, and the two equivalence classes of frames get mapped to the two generators of the homology group if $n>0$ (note that the statement is false for $n=0$ which is why one needs to have better definition of orientations in that case).

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