Odd map implies odd degree with homology and cohomology 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 $H_n(f)(\gamma)=k\,\gamma$.

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.

