This question is both about a specific problem and a soft question about these sorts of problems. I'm reading through Hatcher's textbook and having trouble dealing with signs when it comes to local degree calculations. Here is an example where I am struggling.
Let $f$ be the map $S^n \to \mathbb{R}\mathrm{P}^n$ that is the quotient by antipodal points. I want to show that when $n$ is odd, $f$ sends a generator of $H_n(S^n)$ to twice a generator of $H_n(\mathbb{R}\mathrm{P}^n)$. In order to do that, I can further consider the quotient $q: \mathbb{R}\mathrm{P}^n \to \mathbb{R}\mathrm{P}^n/\mathbb{R}\mathrm{P}^{n-1} \cong S^n$. By the long exact sequence of $(\mathbb{R}\mathrm{P}^n, \mathbb{R}\mathrm{P}^{n-1})$, $q$ induces an isomorphism on homology so it suffices to show that $g :=q \circ f$ has degree $\pm 2$. Let $p$ be the north pole of $S^n$ (i.e. $p = (0, \ldots, 0, 1)$); then $g^{-1}(p)$ consists of $p$ and $-p$. I understand that $g$ is a local homeomorphism around both these points so the local degree is $\pm 1$, but I am not totally sure why the sign of the degrees should be the same.
My vague intuition: let $U_1$ and $U_2$ be neighbourhoods of $p$ and $-p$ that are mapped homeomorphically onto $V$ and let $\alpha$ be the antipodal map. Then $g_{U_1} = g_{U_2} \circ \alpha$, and $\alpha$ sends an oriented generator of $H_n(U_1, U_1 - \{p\})$ to an oriented generator of $H_n(U_2, U_2 - \{p\})$ because $n$ is odd. Therefore $g$ sends generators of both local homology groups to the same element in $H_n(S^n)$. Here I am using the term oriented generator to mean a generator that corresponds to the chosen generator of $H_n(S^n)$ under the isomorphisms $$ H_n(U_1, U_1 - \{p\}) \xrightarrow{i_*} H_n(S^n, S^n-\{p\}) \xleftarrow{j_*} H_n(S^n).$$ But I'd like a more rigorous justification, because I only know that $\alpha$ preserves orientation of a generator of $H_n(S^n)$, and not what it does to local homology groups. Therefore, I tried this: we have the commutative diagram:
The right square commmutes because the underlying maps do, and the left square commutes by naturality of the long exact sequence of homology. The oriented generator of $H_n(U_1, U_1 - \{p\})$ is $e_1 := i_*^{-1}j_*([e])$, where $[e]$ is a chosen generator for $H_n(S^n)$, and similarly for $H_n(U_2, U_2 - \{p\})$. By commutativity of the diagram, $$i_*\alpha_*([e_1]) = \alpha_* i_*([e_1]) = \alpha_* j_*([e]) = j_* \alpha_*([e]) = j_*([e])$$ where in the last equality I have used the fact that $\alpha_*: H_n(S^n) \to H_n(S^n)$ is identity when $n$ is odd. Therefore $\alpha_*([e_1]) = e_2$, i.e., $\alpha$ sends an oriented generator to an oriented generator.
- Have I made any mistakes in this calculation?
- Is there a faster way to rigourously calculate the local degrees in the above? By rigourous, I mean algebraically explicit and not relying on geometric intuition.
- In general I appreciate geometric intuition, but I want to be able to show things algebraically rigourously as well. Is this sort of argument by commutative diagram the usual approach to determine the correct signs of local degrees?