# 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:

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

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:

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

where $\tilde{H}_k$ denotes the reduced homology. Note that for $k \ge 1$, 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.