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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

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You're right that multiplying by a point doesn't change the homology of the space. The problem is that changing the point gives you a different map. The maps relate to each other, but you need to understand how they relate before you can say that the degree does not depend on the point.

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.

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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

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