Additivity of degree Let $\Lambda = \Omega^2$, the product of $\Omega$ two times, be maps of spheres. We know that $\text{deg}(\Lambda) = 2\text{deg}(\Omega)$. How does one show this fact using the formula for the degree? I.e.
$$ \text{deg}(\Omega) = \frac{\int_S \Omega^*(\omega)}{\int_S \omega}$$
I have tried for dimensions $1$ and $2$, but I am unsure how to use the pullback for a cleaner proof. I am mainly interested in the dimension $3$ case, though.
 A: Since you say you are particularly interested in $S^3$, which actually has a group structure, I will consider the case of self-maps of Lie groups (otherwise you haven't made it clear what a "product of maps" $S^k \to S^k$ is supposed to mean in general, since $S^k$ is only a group if $k\in \{0, 1, 3\}$). Caveat: there are some details I'm not 100% sure about because I don't remember how to prove them or where to find a citation.
Let $G$ be a closed, $n$-dimensional Lie group with group operation $\mu\colon G\times G \to G$, let $\omega \in H^n_{dR}(G)$ be non-zero, let $f, g\colon G\to G$ be a smooth functions, and let $f\cdot g\colon G \to G$ be defined by $(f\cdot g)(x) = \mu(f(x), g(x))$. We want to compute
$$deg (f\cdot g) = \frac{\int_G (f\cdot g)^*\omega}{\int_G \omega}  $$
so the first step should be to ask what is the effect of $f\cdot g$ on cohomology. $f\cdot g$ is actually the composition
$$ G \stackrel{f\times g}{\to} G\times G \stackrel{\mu}{\to} G.$$
Claim: $\mu^*(\omega) = \omega\otimes 1 + 1 \otimes \omega$. (I don't remember a proof for this right now, it's possible that it's not actually true.)
From this claim it would follow that $(f\times g)^*\mu^*(\omega) = f^*(\omega)g^*(1) + f^*(1)g^*(\omega) = f^*(\omega) + g^*(\omega)$, so we get
$$deg (f\cdot g) = \frac{\int_G (f\cdot g)^*\omega}{\int_G \omega} = \frac{\int_G (f^*\omega + g^*\omega)}{\int_G \omega}$$ $$ = \frac{\int_G f^*\omega}{\int_G \omega} + \frac{\int_G g^*\omega}{\int_G \omega} = def(f) + deg(g) $$

For maps $f,g\colon S^k \to S^k$ for arbitrary $k$, there is no underlying group structure on $S^k$ to use to define the "product" of $f$ and $g$ (unless $k \in \{0, 1, 3\}$). Alternatively you can use the "concatenation" operation $\ast$ used to define the group structure on $\pi_k(S^k)$ (coming from the fact that every sphere admits a co-group structure) and try to work with a smooth approximation. Then you again need to verify that $(f\ast g)^*(\omega) = f^*(\omega) + g^*(\omega)$, I don't have an argument for this yet.
