Homology for suspension

Statement

In this article, we give the key results relating the homology groups of a topological space and the homology groups of its Suspension (?).

Version for unreduced homology

This states that:

$H_{k + 1} (S X) ≅ H_{k} (X), k \geq 1$

where $H_{k}$ and $H_{k + 1}$ denote the $k^{t h}$ and $(k + 1)^{t h}$ homology groups. The result also holds for homology groups with coefficients.

Further:

$H_{0} (X) ≅ H_{1} (S X) \oplus Z$

and:

$H_{0} (S X) ≅ Z$

Version for reduced homology

This states that:

${\tilde{H}}_{k + 1} (S X) ≅ {\tilde{H}}_{k} (X), k \geq - 1$

where ${\tilde{H}}_{k}$ denotes the reduced homology. Note that for $k \geq 1$ , reduced homology and unreduced homology coincide; for $k = 0$ , the unreduced homology has an extra $Z$ in it. For $k = - 1$ , the right side is the trivial group, giving that ${\tilde{H}}_{0} (S X)$ is trivial, so $S X$ is a path-connected space.

Category-theoretic version

The isomorphisms between the homology groups of a topological space and its suspension are natural isomorphisms between these functors. In particular, if $f : X \to Y$ is a continuous map, then we have an induced continuous map $S f : S X \to S Y$ . There is a commuting diagram relating the homomorphism on $k^{t h}$ reduced homology between $X$ and $Y$ and the homomorphism on $(k + 1)^{t h}$ reduced homology between $S X$ and $S Y$ .

Facts used

Mayer-Vietoris homology sequence

Proof

Recall that $S X$ is obtained by taking $X \times [0, 1]$ (where $[0, 1]$ is the closed unit interval) and then identifying all points in $X \times {1}$ with each other and separately identifying all points in $X \times {0}$ with each other. We will call these two points $p_{1}$ and $p_{0}$ respectively. We consider the following open subsets $U$ and $V$ to use for the Mayer-Vietoris homology sequence:

Subset	Concrete description	Has a strong deformation retraction to ...	More explanation
$U$	$S X ∖ {p_{1}}$	a point (i.e., it is contractible)	$S X ∖ {p_{1}}$ is homeomorphic to the cone space $C X$
$V$	$S X ∖ {p_{0}}$	a point (i.e., it is contractible)	$S X ∖ {p_{0}}$ is homeomorphic to the cone space $C X$
$U \cap V$	$X \times (0, 1)$	$X$	the factor $(0, 1)$ is a contractible space

Proof version with reduced homology

We note that $U$ and $V$ are open subsets and their union is $X$ . Further, because of the strong deformation retraction facts mentioned, all reduced homology groups of $U$ and $V$ are trivial groups and all reduced homology groups of $U \cap V$ are isomorphic to the corresponding reduced homology groups of $X$ .

The original Mayer-Vietoris homology sequence reads:

$\dots \to {\tilde{H}}_{k + 1} (U) \oplus {\tilde{H}}_{k + 1} (V) \to {\tilde{H}}_{k + 1} (S X) \to {\tilde{H}}_{k} (U \cap V) \to {\tilde{H}}_{k} (U) \oplus {\tilde{H}}_{k} (V) \to \dots$

Every third term of this sequence is zero, aand the fragment above simplifies to:

$\dots \to 0 \to {\tilde{H}}_{k + 1} (S X) \to {\tilde{H}}_{k} (X) \to 0 \to \dots$

Since this is a long exact sequence, the map ${\tilde{H}}_{k + 1} (S X) \to {\tilde{H}}_{k} (X)$ is forced to be an isomorphism. This completes the proof.