Prove that for $S^{n}$ (an n-sphere) and given antipodal map A, A is homotopic to the identity on $S^{n}$ Define an antipodal map A such that $$A: S^{n} \rightarrow S^{n}$$ by $A(x_{1},...,x_{n+1})=(-x_{1},...,-x_{n+1}).$ There are other posts on here proving that if n is odd, then A is homotopic to the identity on $S^{n}$ without concerning how the map A is defined. Will the logic change if my antipodal map A is defined as above? I am new to the subject so any input will be greatly helpful. Thank you!

No, in fact the logic works because of some properties:

every map $f:S^n \to S^n$ induced on homology $f_*:H_n(S_n) \to H_n(S_n)$, but since these are just the group $\mathbb Z$, all homomorphisms look like $a \mapsto ka$. We can take $k$ to be the definition of degree. From this, some things follow:

1. $deg(r)$ for any reflection about an axis has degree $-1$
2. $deg(fg)=deg(f)deg(g)$.



Putting these together, and viewing the antipodal map as $n+1$ reflections, we see that $deg(A)=(-1)^{n+1}$.

This is exactly the map the way you define it, and you can see that the degree is $1$ when $n$ is odd, which is the key ingredient in other answers.

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By the way, it would have worked to note that the map is $-I$, where $I$ is the identity matrix. This map is in $SO(n)$, the symmetry group of the sphere when $n$ is odd, and since it is path connected, you can always find a homotopy.