Proof verification in Hatcher's Algebraic Topology, Theorem 3.21

Question

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.

Answer 1

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. $k_1\neq 0$.

Answer 2

We are looking at the map $f : \mathbb{R}\mathrm{P}^{n-1} \to \mathbb{R}\mathrm{P}^{n-1} \times \{y\} \subseteq \mathbb{R}\mathrm{P}^{n-1} \times \mathbb{R}\mathrm{P}^{n-1}$. To understand what this map does to cohomology, it suffices to look at homology (since cohomology is dual). So we have

$$ f_* : H_\bullet(\mathbb{R}\mathrm{P}^{n-1}, \mathbb{Z}_2) \to H_\bullet(\mathbb{R}\mathrm{P}^{n-1} \times \mathbb{R}\mathrm{P}^{n-1}, \mathbb{Z}_2)$$

and it takes a class $\omega \in H_\bullet(\mathbb{R}\mathrm{P}^{n-1})$ to $\omega \times \{y\}$. In particular, $f_*$ takes the dual of $\alpha$ to the dual of $\alpha_1$ where the dual of $\alpha$ is any hyperplane $L \cong \mathbb{R}\mathrm{P}^{n-2} \subset \mathbb{R}\mathrm{P}^{n-1}$ and the dual of $\alpha_1$ is $L \times \{y\}$. This shows that $f^*(\alpha_1) = \alpha$ and $f^*(\alpha_2) = 0$ since $f_*$ doesn't map anything to the dual of $\alpha_2$.

To bring our understanding full-circle, let's understand the map $$f^* : H^i(\mathbb{R}\mathrm{P}^{n-1} \times \mathbb{R}\mathrm{P}^{n-1}, \mathbb{Z}_2) \to H^i(\mathbb{R}\mathrm{P}^{n-1}, \mathbb{Z}_2)).$$ This takes a homomorphism $\phi : H_i(\mathbb{R}\mathrm{P}^{n-1} \times \mathbb{R}\mathrm{P}^{n-1}, \mathbb{Z}_2) \to \mathbb{Z}_2$ to the map $f^*\phi : H_i(\mathbb{R}\mathrm{P}^{n-1}, \mathbb{Z}_2) \to \mathbb{Z}_2$ where

$$ f^*\phi(\omega) = \phi(\omega \times \{y\}).$$

Now let us apply this to the generators of $H^*(\mathbb{R}\mathrm{P}^{n-1} \times \mathbb{R}\mathrm{P}^{n-1}, \mathbb{Z}_2)$. If we fix a hyperplane $L \subset \mathbb{R}\mathrm{P}^{n-1}$, then we can take $\alpha$ to be the homomorphism that maps $L \mapsto 1$. Likewise $\alpha_1 : L \times \{y\} \to 1$ and $\alpha_2 : \{x\} \times L \mapsto 1$. Actually we can describe $\alpha_1, \alpha_2$ better. If $\omega_1 \times \omega_2 \in H_{n-2}$, meaning $\dim \omega_1 + \dim \omega_2 = n - 2$, then

$$ \alpha_1(\omega_1 \times \omega_2) = \begin{cases} 1 & \text{if } \dim \omega_2 = 0 \text{, i.e. } \omega_1 \times \omega_2 \simeq L \times \{y\}, \\ 0 & \text{if } \dim \omega_2 > 0. \end{cases} $$ This is because the cells $\omega_1 \times \omega_2$ with $\dim \omega_2 = 0, 1, 2,\dots,n-2$ for a basis for $H_{n-2}$ and $\alpha_1$ is the dual basis (so it must take the other basis elements to $0$). Likewise, $\alpha_2$ is the dual basis element to $\dim \omega_1 = 0$, i.e. to $\{x\} \times L$.

Finally, we can compute $f^*\alpha_1(L) = \alpha_1(L \times \{y\}) = 1$. Thus $f^*\alpha_1 = \alpha$ because $\alpha$ is exactly the map that takes $L$ to $1$. On the other hand, $f^*\alpha_2(L) = \alpha_2(L \times \{y\}) = 0$ by the previous paragraph. Thus $f^*\alpha_2 = 0$.

Answer 3

Brief aside: In my opinion, Hatcher is an excellent source for examples, and an excellent source for the intuition about some things. I do indeed really like chapters 2 and 3 for these exact reasons. Having said that, I have never found it to be a useful reference for proofs. It may be fruitful when considering technicalities of things like the above, to look at other sources - if for no other reason than getting another opinion.

Anyway, I think that the point is that the inclusion is onto the first factor.

We wish to see what the map is in cohomology which is induced by the inclusion $$i:\Bbb RP^{n-1} \hookrightarrow \Bbb RP^{n-1} \times \Bbb RP^{n-1}$$ (onto the first factor). In particular, for $$i^*:H^*(\Bbb RP^{n-1} \times \Bbb RP^{n-1};\Bbb Z_2 ) \rightarrow H^*(\Bbb RP^{n-1} ;\Bbb Z_2),$$ $$((\alpha_1,\alpha_2)) \mapsto k \alpha,$$ we wish to find $k$.

We do this by looking at the composition with the projection (off the first factor)

$$p_1: \Bbb RP^{n-1} \times \Bbb RP^{n-1} \rightarrow \Bbb RP^{n-1}$$

and the projection (off the second factor)

$$p_2: \Bbb RP^{n-1} \times \Bbb RP^{n-1} \rightarrow \Bbb RP^{n-1}$$

We have trivially that the composition $p_1 \circ i: \Bbb RP^{n-1} \rightarrow \Bbb RP^{n-1}$ is the identity $1_{\Bbb RP^{n-1}}$, so that the induced map in cohomology of this composition is the identity: $(p_1 \circ i)^* = 1_{H^*(\Bbb RP^{n-1} ;\Bbb Z_2)}$.

However, the composition of the inclusion onto the first factor with the projection off the second factor $p_2 \circ i$ is the constant map at the basepoint. That is, it induces the trivial map in cohomology.

From the above, I think Hatcher's claim follows immediately.

We then use that since $h$ is a homeomorphism when restricted to the first factor (that is, restrics to a homeomorphism when pre-composed with the inclusion $i:\Bbb RP^{n-1} \hookrightarrow \Bbb RP^{n-1} \times \Bbb RP^{n-1}$), the induced map on cohomology cannot be identically zero. Since $k_2$ is zero, thus $k_1$ can't be.