Degree of map $S^{n-1} \times S^{n-1} \to S^{n-1}$ when restricted to one of the factors

We can define the degree, $d$ of a continuous map $f:S^n \to S^n$ through the induced map, $f_*$, in homology: $x \mapsto dx$.

Now consider a map $S^{n-1} \times S^{n-1} \to S^{n-1}$, and let $y \in S^{n-1}$. Let $\alpha$ denote the degree of $g$ restricted to $S^{n-1} \times y$ and $\beta$ the degree of $g$ restricted to $y \times S^{n-1}$. Show that $\alpha$ and $\beta$ are independent of the choice of $y$.

I am not really sure what I need to prove here! If the degree $\alpha$ is simply given by the induced map $H_{n-1}(S^{n-1} \times y) \to H_{n-1}(S^{n-1})$, then there is really nothing to prove since multiplying by a point $y$ does not change the homology.

I am sure that is not exactly what is required. So, any hints for what to do in this question?

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A strange problem: $S^n$ is a connected manifold, so for all $x, y \in S^n$ there is a homeomorphism $f: S^n \to S^n$ such that $f(x) = y$, and thus the proof should be trivial. –  Alexei Averchenko Aug 15 '11 at 8:06

Here is a hint: if $f,g:X\to Y$ are homotopic, then the induced maps on homology are equal, i.e., $f_*=g_*$. Now use that $S^{n-1}$ is path connected.
Thanks. Is the essential argument that the choice of $y$ does not matter because there is always a path between two choices of $y$? –  Juan S Aug 15 '11 at 7:32
@Qwirk Yes, and the path gives rise to a map of the cylinder $S^{n-1}\times I \to S^{n-1}\times S^{n-1}$. The cylinder gives the desired homotopy. Unfortunately, we cannot use this to say that $\alpha=\beta$ because the embedded spheres are not homotopic: $\pi_n(S^n\times S^n)\cong \pi_n(S^n)\times \pi_n(S^n)\cong \mathbb Z^2$, and the cross sections in the problem correspond to two generators of the homotopy group. –  Aaron Aug 15 '11 at 8:08