I'm trying to prove that the mapping $r_i\colon\mathbb{R}^n\rightarrow\mathbb{R}^n$, $r_i(x_1,\dots,x_n)=(x_1,\dots,-x_i,\dots,x_n)$ induces multiplying by (-1) in $H_{n-1}(S^{n-1})$.
We see that $r_i(S^{n-1})=S^{n-1}$. I wanted to show that n-chain $z\in C_n(S^{n-1})$ is mapped to $-z$, but I don't know how and if it is so (then using a differential would finish the proof). Of course $r_i$ is a homeomorphism and homotopy equivalence. Hence, $H_{n-1}(r_i)\colon H_{n-1}(S^{n-1})\rightarrow H_{n-1}(S^{n-1})$ is a homomorphism, but I don't know how to connect these two mappings.
Any hints?
Edit: but hey! isn't that ok? This reflection maps chain z to z', and then rotation around any other axis by $\pi$ maps it on a $-z$. This is a homotopy equivalence, so doesn't change homology group, and then we can use the differential, which would finish the proof. Is that a good idea? Please, feel free to comment.
Thanks in advance
The idea captures some nice intuition but if you want to be formal, then in singular homology the $\pi$-rotation doesn't map it to $"-z"$: formally those are different singular simplices (although homologous).
The simplest way to solve this is to build a simplicial complex model on the sphere and approximate $r$ via a simplicial map that is homotopic to $r$. Then you just follow the definitions.
A more abstract way is by induction and Mayer-Vietoris: if $S$ is a neighborhood of the hemisphere $\{j\geq 0\}$ for some $j\neq i$ and $N$ a neighborhood of the opposite hemisphere, $S\cap N$ is an $(n-2)$-sphere up to homotopy equivalence, both $N$ and $S$ are fixed by $r$ and you have the diagram $$ \begin{array}{ccc} H_{n-1}(S^{n-1})=H_{n-1}(N\cup S)&\simeq & H_{n-2}(S\cap N)\simeq H_{n-2}(S^{n-2})\\ \downarrow_{r} & & \downarrow_r\\ H_{n-1}(S^{n-1})=H_{n-1}(N\cup S)&\simeq & H_{n-2}(S\cap N)\simeq H_{n-2}(S^{n-2})\\ \end{array} $$ Of course, the restriction of $r$ to the "equator" $R\cap S$ has the same form and sends a generator to its inverse, so the same is true for $H_{n-1}$. For the $S^0$-shpere, you again can do it directly from the definition.
Yet another insight is the following: $$H_{n-1}(S^{n-1})\simeq H_{n-1}(R^n\setminus\{0\})\simeq H_n(R^n,R^n\setminus\{0\})$$ by the deformation retraction and long exact sequence: but the homology $H_n(R^n, R^n\setminus\{0\})$ is an "orientation" (see Hatcher, 233ff) which is changed by $r$.