I'm trying to show that if $f : S^n \to S^n$ is a map with degree $d$, then the induced map $f^*$ on any (reduced, ordinary) cohomology theory is multiplication by $d$.
Some background:
Cohomology theories:
A reduced extraordinary cohomology theory is a sequence of contravariant functors $\tilde E^n$ (for $n \in \mathbb Z$) from CW complexes to abelian groups, with natural transformations $\delta : \tilde E^n(A) \to \tilde E^{n+1}(X/A)$ for CW pairs $(X,A)$ satisfying the following axioms:
- Homotopy: if $f, g : X \to Y$ are homotopic then the induced maps $f^*,g^* : \tilde E^n(Y) \to \tilde E^n(X)$ are equal.
- Exactness: for each CW pair $(X,A)$ there is a long exact sequence $$... \xrightarrow{\delta} \tilde E^n(X/A) \xrightarrow{q^*} \tilde E^n(X) \xrightarrow{i^*} \tilde E^n(A) \xrightarrow{\delta} \tilde E^{n+1}(X/A) \xrightarrow{q^*} ...$$ where $i$ is the inclusion map $A \to X$ and $q$ is the quotient map $X \to X/A$.
- Additivity: For a wedge sum $X = \vee_\alpha X_\alpha$ with inclusions $i_\alpha : X_\alpha \hookrightarrow X$, the product map $$\prod_\alpha i_\alpha^* : \widetilde E^n(X) \xrightarrow{\cong} \prod_\alpha \widetilde E^n(X_\alpha)$$ is an isomorphism for all $n$.
A reduced (ordinary) cohomology theory is a reduced extraordinary cohomology theory satisfying an additional axiom
- Dimension: The cohomology groups $\widetilde E^n(*)$ are 0 for all $n$, where $*$ is the space consisting of a point.
Degree
Let $f : S^n \to S^n$. The degree of $f$ is the integer $[f] \in \pi_n(S^n) \cong \mathbb Z$.
See corollary 4.25 of Hatcher for proof $\pi_n(S^n) \cong \mathbb Z$. Corollary 4.25 also tells us that this definition of degree agrees with the usual definition (where the degree of $f : S^n \to S^n$ is the integer $d$ in the induced map on cohomology $f^* : H^n(S^n) \cong \mathbb Z \to H^n(S^n) \cong \mathbb Z : \alpha \mapsto d\alpha$).
My proof so far:
Proposition. If $f : S^n \to S^n$ has degree $d$, then the induced map $f^*$ on any reduced cohomology theory is multiplication by $d$.
Proof: For $d \ge 1$, $f$ is homotopic to $$g_d : S^n \xrightarrow{\mathrm{pinch}} \vee_d S^n \xrightarrow{\vee \mathrm{id}} S^n$$ (since we know $g_d$ has degree $d$).
If $E$ is any reduced cohomology theory, then $$ E^k(S^n) \xrightarrow{(\vee \mathrm{id})^*} E^k\left(\vee_d S^n \right) \cong \prod_d E^k(S^n) \xrightarrow{(\mathrm{pinch})^*} E^k(S^n) $$
where $(\vee \mathrm{id})^* : x \mapsto (x,...,x)$ and $(\mathrm{pinch})^*: (x_1,...,x_d) \mapsto \sum_i x_i$.
(Why is $(\mathrm{pinch})^*(x_1,...,x_d) = \sum_i x_i$? When $d = 1$, we know that the pinch map is homotopic to the identity. It follows that $(\mathrm{pinch})^*(0,...,0,x_i,0,...,0) = x_i$ and there is only one homomorphism satisfying this property.) Thus, the composition is multiplication by $d$.
For $d = 0$, $f$ is null-homotopic, so $f^*$ must be multiplication by 0. (Why is the constant map multiplication by 0? It factors through $*$, which has reduced cohomology groups all 0.)
For $d \le -1$, $f$ is homotopic to a reflection $r$ composed with $g_{-d}$, since we know that $r$ has degree $-1$. (By a reflection, we mean a map that fixes a great circle $S^{n-1}$ and interchanges the corresponding hemispheres.)
I want to show that $r^*$ is multiplication by $-1$, but I don't know how. This would then complete the proof.
