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 fact about::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) uses::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:

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.