First, this is the link of the book, for convenience: https://pi.math.cornell.edu/~hatcher/AT/AT.pdf#page=232
I was reading the proof of Theorem 3.21, and I have some problems verifying the proof. The (front part of the) summarized proof is as follows:
Theorem 3.21. If $\Bbb R^n$ has a structure of a division algebra over $\Bbb R$, then $n$ must be a power of $2$.
Proof. Since the multiplication map $\Bbb R^n \times \Bbb R^n \to \Bbb R^n$ is bilinear, it is continuous, and it induces a map $h : \Bbb RP^{n-1} \times \Bbb RP^{n-1} \to \Bbb RP^{n-1}$ which is a homeomorphism when restricted to each subspace $\Bbb RP^{n-1} \times \{y \}$ and $\{x \} \times \Bbb RP^{n-1}$. Consider the induced ring homomorphism $$h^* : H^*(\Bbb RP^{n-1} ;\Bbb Z_2) \cong \Bbb Z_2[\alpha]/(\alpha^n) \to H^*(\Bbb RP^{n-1} \times \Bbb RP^{n-1};\Bbb Z_2 ) \cong \Bbb Z_2[\alpha_1,\alpha_2]/(\alpha_1^n,\alpha_2^n) ,$$ where $$H^*(\Bbb RP^{n-1} \times \Bbb RP^{n-1};\Bbb Z_2 ) \cong \Bbb Z_2[\alpha_1,\alpha_2]/(\alpha_1^n,\alpha_2^n)$$ holds by the Kunneth formula.
Letting $h^*(\alpha)=k_1 \alpha_1+ k_2 \alpha_2$, the inclusion $\Bbb RP^{n-1} \to \Bbb RP^{n-1} \times \Bbb RP^{n-1}$ onto the first factor sends $\alpha_1$ to $\alpha$ and $\alpha_2$ to $0$, as one sees by composing with the projections of $\Bbb RP^{n-1} \times \Bbb RP^{n-1}$ onto its two factors. Then the fact that $h$ restricts to a homeomorphism on the first factor implies $k_1$ is nonzero.
I understood the first paragraph, but I cannot understand the second paragraph at all. Why does the inclusion sends $\alpha_1$ to $\alpha$ and $\alpha_2$ to $0$? Also, how does the next sentence follow? Thanks in advance.
Best Answer
Let $X,Y$ be nonempty path-connected spaces, $\alpha \in H^p(X;k), \beta \in H^q(Y;k)$ for some field $k$.
Then you get cohomology classes $\alpha \otimes 1 \in H^p(X;k)\otimes H^0(Y;k) \subset H^p(X\times Y;k)$ and $1\otimes \beta \in H^0(X;k)\otimes H^q(Y;k) \subset H^q(X\times Y;k)$.
I claim that the inclusion $i:X\to X\times Y$ at any basepoint of $Y$ sends $\alpha\otimes 1$ to $\alpha$ and $1\otimes \beta$ to $0$. If you apply this in your particular situation with $X=Y= \mathbb RP^{n-1}$ , $\alpha_1$ and $\alpha_2$, you get the desired result.
To prove the claim, you can use the projection $p:X\times Y \to X$ and notice that $p\circ i = id_X$, so that when you look at cohomology, $i^*\circ p^* = id$ on cohomology. In particlar, $i^*(\alpha\otimes 1) = i^*(p^*(\alpha)) = \alpha$. Here, I used $p^*\alpha = \alpha\otimes 1$, but this is essentially part of the Künneth theorem.
Similarly, we may use the projection $q: X\times Y\to Y$ and notice that $q\circ i$ is a constant map, hence it induces $0$ on cohomology. It follows that $0= i^*(q^*(\beta)) = i^*(1\otimes \beta)$; using $q^*\beta = 1\otimes \beta$, which again is essentially part of the Künneth theorem.
This solces the question of "why does the inclusion act like that on $\alpha_1, \alpha_2$ ?"
Then for the second question (how the next sentence follows), consider then $i^*h^*\alpha = i^*(k_1\alpha_1+k_2\alpha_2 ) = k_1i^*\alpha_1 + k_2i^*\alpha_2 = k_1\alpha$ by the previous computation, and $h\circ i$ is a homeomorphism (by what has been said before).
Therefore $i^*h^* = (h\circ i)^*$ is an isomorphism on cohomology, and since $\alpha\neq 0$, it must be that $i^*h^*\alpha \neq 0$, i.e. $k_1\neq 0$.