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Let $p_2:\mathbb S^1 \to \mathbb S^1$ be the two-sheeted covering map $p(z)=z^2$.If $f$ is odd($f(-z)=-f(z)$),show that there exists a continuous map $g:\mathbb S^1 \to \mathbb S^1$ such that $\deg f=\deg g$ and the following diagram commutes:$p_2 \circ f=g \circ p_2$

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What are your thoughts on the matter? –  t.b. May 29 '12 at 14:00
    
@t.b. I know two continuous map are homotopic iff their degree are equal.So we could find a continuous map $g$ satisfies $\deg f=\deg g$ and $p_2 \circ f=g \circ p_2$.So f and g are homotopic and $p_2 \circ f$ and $g \circ p_2$ are homotopic.But I don't know how to use f is odd and how to show $p_2 \circ f$ and $g \circ p_2$ are equal. –  Jiangnan Yu May 29 '12 at 14:05

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