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I have trouble in following one proof in Hatcher's Algebraic topology. On Page 210, Section 3.2 cup product, in Example 3.11 ($n$-Torus) it is claimed that the sequence $$ 0\to H^n(I\times Y; R)\to H^n(\partial I\times Y;R)\overset{\delta}{\to}H^{n+1}(I\times Y,\partial I\times Y;R)\to 0 $$ is exact.

I can't see how to get the zeros at the two ends (i.e. injectivity and surjectivity).

Also, I can't see why the map $$ H^{n+1}(Y;R)\times H^n(Y;R)\to H^{n+1}(S^1\times Y;R) $$ on the next page is an isomorphism.

Can someone help me? Thanks.

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    $\begingroup$ Very coincidental, I was just struggling with this very proof yesterday. $\endgroup$
    – Eric Auld
    Aug 2, 2013 at 22:40

1 Answer 1

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Consider the composition $$\{0\}\times Y\hookrightarrow\partial I\times Y\hookrightarrow I\times Y$$ When we apply the cohomology functor, we get: $$H^n(\{0\}\times Y;R)\twoheadleftarrow H^n(\partial I\times Y;R)\hookleftarrow H^n(I\times Y;R)$$ since the homomorphism induced by the homotopy equivalence $\{0\}\times Y\hookrightarrow I\times Y$ is bijective.

These injections appear in the long exact sequence of the pair $(I\times Y,\partial I\times Y)$

$$...\to H^n(I\times Y;R)\hookrightarrow H^n(\partial I\times Y;R)\to H^n(I\times Y,\partial I\times Y;R)\to H^{n+1}(I\times Y;R)\hookrightarrow...$$

Now, it is a property of exact sequences that a map is surjective if and only if the map two steps further is injective. Hence we get a surjection: $$H^n(\partial I\times Y;R)\twoheadrightarrow H^n(I\times Y,\partial I\times Y;R)$$

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