I have problem with such a proof, namely, for $f$ - odd function we get the local degree $d(f)$ is odd. Generaly, we start with $f$ - $\mathcal{C}^{\infty}$, odd function, then there is a construction concidered a hyperplanes (point is, that we can take such a hyperplane, that non of regular points $\left\{p_1,\ldots p_r\right\}$ = preimages of a regular value $q\in\mathcal{S}^n$. We know also that $$d(f)=r \ (mod \ 2).$$
Now we define hemispheres, and two maps: $g$ and $\rho$ - they are acutally on the picture. As you see, $\rho$ is odd, and $g$ is as below.
My question is, how from below diagram we can deduce eqaulity (6): $$d(f)=\underline{+}d(\rho\circ g|_{S^n})$$
By definition, $d(g)$ is the integer $d$ such that the leftmost map in the diagram $H(g):H(S^n)\to H(S^n)$ is given by multiplication by $d$. Similarly, $d(\rho\circ g|_{S^{n-1}})$ is the integer $d$ such that the rightmost map $\tilde{H}(\rho\circ g):\tilde{H}(S^{n-1})\to \tilde{H}(S^{n-1})$ is multiplication by $d$. Now if you choose generators of $H(S^n)$ and $\tilde{H}(S^{n-1})$ and so identify them with $\mathbb{Z}$ and compose all the horizontal maps together to get just isomorphisms from the left all the way to the right, you get a commutative diagram $$\require{AMScd} \begin{CD} \mathbb{Z} @>{}>> \mathbb{Z}\\ @V{d(g)}VV @VV{d(\rho\circ g|_{S^{n-1}})}V \\ \mathbb{Z} @>{}>> \mathbb{Z} \end{CD}$$ where the top and bottom maps are both isomorphisms. An isomorphism $\mathbb{Z}\to\mathbb{Z}$ must be multiplication by $\pm 1$. So commutativity of this square says that $(\pm 1)d(g)=(\pm 1)d(\rho\circ g|_{S^{n-1}})$ for some choice of signs, or $d(g)=\pm d(\rho\circ g|_{S^{n-1}})$.