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Suppose $n$ is odd. Let $f:S^n\longrightarrow S^n$ be an odd function. Then it induces a map $g:P^n\longrightarrow P^n$ such that the diagram $$ \begin{array}[c]{ccc} S^n&\stackrel{f}{\rightarrow}&S^n\\ \downarrow\scriptstyle{p}&&\downarrow\scriptstyle{p}\\ P^n&\stackrel{g}{\rightarrow}&P^n \end{array} $$ We can also guarantee that the antipodal map has degree 1.

Can I deduce from the above that the degree of $f$ is odd? I have tried using $H_n(f)$ with coefficients on $\mathbb{Z}$ and on $\mathbb{Z}_2$, but I have not been able to prove it.

My definition of degree is, given $\gamma\in H_n(S^n)$ a generator, then $\deg(f)=k$ with $k$ such that $H_n(f)(\gamma)=k\,\gamma$.

user337921
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1 Answers1

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It is enough to show that $H_n(g):H_n(\Bbb RP^n)\to H_n(\Bbb RP^n)$ is multiplication by an odd integer because $2\deg(f)=\deg(p\circ f)=\deg(g\circ p)=2\deg(g)$.

By the UCT, this is equivalent to showing that $$H_n(g;\Bbb Z/2):H_n(\Bbb RP^n;\Bbb Z/2)\to H_n(\Bbb RP^n; \Bbb Z/2)$$ is an isomorphism, which in turn is equivalent to the analogous statement for $$H^n(g;\Bbb Z/2):H^n(\Bbb RP^n;\Bbb Z/2)\to H^n(\Bbb RP^n; \Bbb Z/2).$$

You can prove this as follows:

  • From the fact that $f$ is odd, prove that $\pi_1(g)$ is an isomorphism. (Lift the generating loop $\alpha:I\to \Bbb RP^n$ to a path in $S^n$ and use commutativity of the diagram from your question)
  • From the ring structure on $\Bbb RP^n$ conclude that $H^n(g; \Bbb Z/2)$ is an isomorphism.
iwriteonbananas
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  • Two questions about your answer. First, I don't understand why we can lift $\alpha$ to a loop in $S^n$. I get that we get a path in $S^n$, but not a loop. Second, after 1 I can deduce that $H_1$ is an isomorphism. How do I dedude anything about $H^n$. Using duality? – user337921 May 13 '16 at 20:57
  • You're right, it's won't be a loop but only a path, @user337921. When $H_1(g;\Bbb Z_2)$ is an isomorphism, then so is $H^1(g;\Bbb Z_2)$. Now you can exploit the ring structure $H^*(\Bbb RP^n; \Bbb Z_2)\cong \Bbb Z_2[a]/(a^{n+1})$ to conclude that $H^n(g;\Bbb Z_2)$ is an isomorphism. Do you see how that works? Are you familiar with the cohomology ring structure? – iwriteonbananas May 14 '16 at 09:43
  • Yes, I am familiar with the cohomology ring. My problem is proving that $\pi_1(g)$ is isomorphism, $H_1(g)$ isomorphism implies $H_1(g,\mathbb{Z}_2)$ isomorphism, and then $H_1(g,\mathbb{Z}_2)$ implies $H^1(g,\mathbb{Z}_2)$. – user337921 May 14 '16 at 14:47