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Relative homology of an h-cobordism vanishes at both ends

Statement

Let $(W; M_{0}, M_{1})$ be an h-cobordism (h-Cobordism) and let $G$ be any abelian coefficient group. Then $H_{k} (W, M_{0}; G) = 0$ and $H_{k} (W, M_{1}; G) = 0$ for every $k \geq 0$ ; equivalently, each inclusion $M_{i} ↪ W$ induces an isomorphism $H_{k} (M_{i}; G) \to H_{k} (W; G)$ for all $k \geq 0$ . In particular $H_{k} (W, M_{i}; Z) = 0$ for both ends. No orientability, finite generation or simple connectivity hypothesis is used, and the conclusion is symmetric in the two ends.

Facts & Assumptions

Given: An h-cobordism $(W; M_{0}, M_{1})$ and an abelian group $G$ .

[F1]

Both face inclusions of an h-cobordism are homotopy equivalences: the triad $(W; M_{0}, M_{1})$ has $\dim W = n + 1 \geq 2$ and both $M_{0} ↪ W$ and $M_{1} ↪ W$ are homotopy equivalences; the definition is symmetric in the two faces (h-Cobordism).

[F2]

If $f : X \to Y$ is a homotopy equivalence, then for every $n \geq 0$ and every abelian group $G$ the induced map $H_{n} (f_{#}) : H_{n}^{s i n g} (X; G) \to H_{n}^{s i n g} (Y; G)$ is an isomorphism (Homotopy equivalences induce isomorphisms on singular homology).

[F3]

For every subspace $A \subseteq X$ the singular homology groups of the pair form the long exact sequence $\dots \to H_{k} (A; G) \to H_{k} (X; G) \to H_{k} (X, A; G) \overset{δ}{\to} H_{k - 1} (A; G) \to H_{k - 1} (X; G) \to \dots$ (the display is printed for $n$ and degree $n - 1$ in the source), and $H_{k} (X, A; G)$ denotes the relative singular homology group in degree $k$ (Long exact sequence of a pair, Relative singular homology).

Proof

technique · direct

1.1F1F2given

By [F1] the inclusion $ι_{0} : M_{0} ↪ W$ is a homotopy equivalence, so by [F2] the induced map $H_{k} (ι_{0}; G) : H_{k} (M_{0}; G) \to H_{k} (W; G)$ is an isomorphism for every $k \geq 0$ , in particular surjective; the same applies to $ι_{1} : M_{1} ↪ W$ .

2.1F3step 1.1

Fix $k \geq 0$ and read the exact sequence of [F3] for the pair $(W, M_{0})$ around $H_{k} (W, M_{0}; G)$ , namely $H_{k} (M_{0}; G) \overset{H_{k} (ι_{0}; G)}{\to} H_{k} (W; G) \to H_{k} (W, M_{0}; G) \overset{δ}{\to} H_{k - 1} (M_{0}; G) \overset{H_{k - 1} (ι_{0}; G)}{\to} H_{k - 1} (W; G)$ . Surjectivity of $H_{k} (ι_{0}; G)$ together with exactness at $H_{k} (W; G)$ makes $H_{k} (W; G) \to H_{k} (W, M_{0}; G)$ the zero map; injectivity of $H_{k - 1} (ι_{0}; G)$ together with exactness at $H_{k - 1} (M_{0}; G)$ makes $δ$ the zero map (for $k = 0$ the group $H_{- 1} (M_{0}; G)$ is zero, so $δ$ is automatically zero). The image of the zero map $H_{k} (W; G) \to H_{k} (W, M_{0}; G)$ is $0$ , so exactness at $H_{k} (W, M_{0}; G)$ gives $\ker δ = 0$ , and since $δ = 0$ this says $H_{k} (W, M_{0}; G) = 0$ .

3.1F1F2F3step 1.1step 2.1∎

Interchanging the roles of $M_{0}$ and $M_{1}$ , which is legitimate for an h-cobordism by [F1], the argument of step 2.1 gives $H_{k} (W, M_{1}; G) = 0$ for every $k \geq 0$ . By step 1.1 each inclusion induces an isomorphism in every degree, and specialising $G = Z$ gives $H_{k} (W, M_{0}; Z) = H_{k} (W, M_{1}; Z) = 0$ .

Depends on

Used by

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Sources

John Milnor, Lectures on the h-Cobordism Theorem (notes by L. Siebenmann and J. Sondow, Princeton University Press 1965; scanned edition with searchable text layer) (standard reference, not scraped)
Wolfgang Lück, A Basic Introduction to Surgery Theory (ICTP lecture notes, 27 October 2004; complete author text) (standard reference, not scraped)