4
$\begingroup$

On pg 210 and 211(section 3.2, example 3.11), Hatcher attempts to prove that $H^{k}(T^{n},R)$ has basis the cup products $\alpha_{i_{1}} \smile \cdots \smile \alpha_{i_{k}}$ where $i_{1}< \cdots < i_{k}$ and $T^{n} $ is the n-Torus. He first shows that if $\alpha \in H^{1}(I,\partial I;R)$ is a generator, the map given by the external cross product:

$H^{n}(Y;R) \to H^{n+1}(I \times Y,\partial I \times Y;R)$ given by $\beta \mapsto \alpha \times \beta$ is an isomorphism. To do this, he establishes the simple fact of naturality with respect to coboundary maps in the long exact sequence of relative cohomology. He mentions the following split short exact sequence obtained for the pair $(I \times Y,\partial I \times Y)$

$0 \rightarrow H^{n}(I \times Y;R) \rightarrow H^{n}(\partial I \times Y;R) \xrightarrow{\delta} H^{n+1}(I \times Y,\partial I \times Y;R) \rightarrow 0$. I did some calculations to verify that it is indeed split. However, I can't understand any of the explanation following that

Hatcher

How is $\delta$ an isomorphism on the restriction and how exactly is $\beta \mapsto \alpha \times \beta$ an isomorphism? Of course, the entire problem becomes much easier using the Kunneth formulas but I'd like to understand this proof too.

$\endgroup$
1
$\begingroup$

I will assume that all cohomology groups have coefficients in $R$ and omit it from the notation. Let $1_0$ the $0$-cocycle defined as in the example, and let $1_1$ be defined analogously. Then it is not hard to show that every cocycle in $H^n(\partial I \times Y)$ can be written uniquely as a sum $1_0 \times \beta_0 + 1_1 \times \beta_1$, where $\beta_0, \beta_1 \in H^n(Y)$.

Let's start by proving that $\delta$ is injective when restricted to elements of the form $1_0 \times \beta$ for $\beta \in H^n(Y)$. Suppose that $\delta(1_0 \times \beta) = \delta(1_0) \times \beta = 0$. Let $\sigma : [v_1, \dots, v_{n+1}] \to Y$ be any singular $n$-simplex in $Y$. There is a singular $(n+1)$-simplex $\tilde{\sigma} : [v_0, \dots, v_{n+1}] \to I \times Y$ such that $p \circ \tilde{\sigma}|_{[v_1, \dots, v_{n+1}]} = \sigma$ where $p : I \times Y \to Y$ is the projection, and such that $\tilde{\sigma}(v_0) \in \{1\} \times Y$ and $\tilde{\sigma}([v_1, \dots, v_{n+1}]) \subseteq \{0\} \times Y$. Let $q : I \times Y \to I$ be the other projection. Then $$ 0 = (\delta(1_0) \times \beta)(\tilde\sigma) = (q^*\delta(1_0) \smile p^*\beta)(\tilde\sigma) = q^*\delta(1_0)(\tilde\sigma|_{[v_0,v_1]}) \cdot p^*\beta(\tilde\sigma|_{[v_1, \dots, v_{n+1}]}) = \beta(\sigma). $$ Therefore $\beta = 0$, and $\delta$ is injective.

To prove surjectivity, first note that $\delta(1_1) = -\delta(1_0)$. To see this, we extend $1_0 + 1_1$ to all of $I$ by setting equal to $1$ everywhere. Then $\delta(1_0 + 1_1) = 0$, proving the claim. Since the sequence at the bottom of page 210 is exact, we have that $\delta$ is surjective when its domain is taken to be all of $H^n(\partial I \times Y)$. Hence, for every $\psi \in H^{n+1}(I \times Y, \partial I \times Y)$, we have some $\beta_0, \beta_1$ such that $$ \psi = \delta(1_0 \times \beta_0 + 1_1 \times \beta_1) = \delta(1_0) \times \beta_0 + \delta(1_1) \times \beta_1 = \delta(1_0) \times (\beta_0 - \beta_1) = \delta(1 \times (\beta_0 - \beta_1). $$ this concludes the proof of surjectivity, and therefore that $\delta$ is an isomorphism when restricted to a copy of $H^n(Y)$.

It follows from the above that $\beta \mapsto \delta(1_0) \times \beta$ is an isomorphism. Since for any unit $r \in R$, $\delta(1_0) \times \beta \mapsto (r\delta(1_0)) \times \beta$ is an isomorphism, we conclude that $\beta \mapsto \alpha \times \beta$ is an isomorphism for any $\alpha$ generating $H^1(I, \partial I)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.