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:

$H_{k+1}(SX)\cong H_{k}(X),k\geq 1$

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

Further:

$H_{0}(X)\cong H_{1}(SX)\oplus \mathbb {Z}$

and:

$H_{0}(SX)\cong \mathbb {Z}$

Version for reduced homology

This states that:

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

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 $\mathbb {Z}$ in it. For $k=-1$, the right side is the trivial group, giving that ${\tilde {H}}_{0}(SX)$ is trivial, so $SX$ 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 $Sf:SX\to SY$. There is a commuting diagram relating the homomorphism on $k^{th}$ reduced homology between $X$ and $Y$ and the homomorphism on $(k+1)^{th}$ reduced homology between $SX$ and $SY$.

Facts used

1. Mayer-Vietoris homology sequence

Proof

Recall that $SX$ 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
$U$	$SX\setminus \{p_{1}\}$	a point (i.e., it is contractible)	$SX\setminus \{p_{1}\}$ is homeomorphic to the cone space $CX$
$V$	$SX\setminus \{p_{0}\}$	a point (i.e., it is contractible)	$SX\setminus \{p_{0}\}$ is homeomorphic to the cone space $CX$
$U\cap V$	$X\times (0,1)$	$X$	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:

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

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

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

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