# Reduced Homology on unreduced suspension

I want to establish the result for any homology theory, there is a natural isomorphism which gets us $\tilde{H}_{p+1}(\Sigma X) \cong \tilde{H}_p(X)$. Most places use the Mayer-Vietoris sequence to get this, but since I haven't covered that yet, I would like to do it using the Eilenberg-Steenrod axioms for homology.

Edit: Note that I'm defining $\Sigma X$ as the space where I shrink $X \times \{0\}$ and $X \times\{1\}$ to points. Wikipedia calls this $SX$.

Is there any difference between the reduced and unreduced suspension? This is related with this arguing here: https://mathoverflow.net/questions/107430/does-the-reduced-mapping-cylinder-have-the-same-homotopy-type-of-unreduced-mappin

The reduced cone has the same homotopy type of the unreduced one, when we are talking about well based spaces. The problem is if the inclusion $X\to CX$, in which $CX$ is the reduced cone, is a unbased cofibration. (I don't think it is in general: only if it is a well pointed space). Assuming it is well based space, you don't have to worry if it is the reduced suspension or the unreduced one.

Back to the problem: We are only considering unreduced constructions, as I guess you want. This is consequence of the excision and the exact sequence. I guess you are talking about Eilenberg-Steenrod axioms for unreduced homology, since the Eilenberg-Steenrod axioms for the reduced homology assumes the suspension isomorphism as an axiom.

$(\sum X , CX , CX)$ is a excisive triad. By the axiom, you get that $(CX,X)\to (\sum X, CX)$ induces isomorphisms between the homology groups. Well, now, using the exact sequence of the pair $(CX, X)$, we can prove that $H_q(X, \ast )$ is isomorphic to $H_{q+1}(CX,X)$.

TO prove this last statement, first of all, you have to notice that $H_q(X)\cong H_q(X, \ast)\oplus H_q(\ast )$. And, then, notice that that we get a new exact sequence from this congruence. This new exact sequence is

$\cdots \rightarrow H_q(A, \ast )\to H_ q(X, \ast )\to H_q(X,A)\to H_{q-1}(X, \ast)\rightarrow \cdots$.

In our case, $\cdots \rightarrow H_q(X, \ast )\to H_ q(CX, \ast )\to H_q(CX,X)\to H_{q-1}(X, \ast)\rightarrow \cdots$.

Since $H_ q(CX, \ast )$ is clearly trivial, we conclude that

$H_q(CX,X)\to H_{q-1}(X, \ast)$

is an isomorphism.

So we can conclude that there is an isomorphism $H_q(X, \ast)\to H_ {q+1}(\sum X, CX)$. Since $(\sum X, CX)\equiv (\sum X, \ast)$, the required statement was proven.

Disclaimer: The following is a proof that does not use the Mayer-Vietoris sequence, but neither does it strictly lie within the framework of the Eilenberg-Steenrod axioms for homology (as there is mention of chain complexes), which is being requested for by the OP.

Let $X$ be a (non-empty) topological space. Contracting the base $X \times \{ 0 \}$ of the cone $C X$, we obtain the unreduced suspension $\Sigma X$. In other words, $$\Sigma X \stackrel{\text{homeo}}{\cong} C X / (X \times \{ 0 \}).$$ Consider the short exact sequence of chain complexes $$0 \to {\Delta_{\bullet}}(X \times \{ 0 \}) \to {\Delta_{\bullet}}(C X) \to {\Delta_{\bullet}}(C X,X \times \{ 0 \}) \to 0,$$ which in dimension $-1$ is $$0 \to \mathbb{Z} \stackrel{\text{id}}{\to} \mathbb{Z} \to 0 \to 0.$$ There is a corresponding long exact sequence of reduced homology groups: $$\cdots \to {\tilde{H}_{n + 1}}(C X) \to {\tilde{H}_{n + 1}}(C X,X \times \{ 0 \}) \to {\tilde{H}_{n}}(X \times \{ 0 \}) \to {\tilde{H}_{n}}(C X) \to \cdots.$$ As $C X$ is contractible, we have ${\tilde{H}_{n}}(C X) = 0$ for all $n \in \mathbb{Z}$. Hence, $$\forall n \in \mathbb{Z}: \quad {\tilde{H}_{n + 1}}(C X,X \times \{ 0 \}) \cong {\tilde{H}_{n}}(X \times \{ 0 \}).$$ This, however, yields \begin{align*} \forall n \in \mathbb{Z}: \quad {\tilde{H}_{n + 1}}(\Sigma X) & \cong {\tilde{H}_{n + 1}}(C X / (X \times \{ 0 \})) \quad (\text{As $\Sigma X \stackrel{\text{homeo}}{\cong} C X / (X \times \{ 0 \})$.}) \\ & \cong {H_{n + 1}}(C X,X \times \{ 0 \}) \quad (\text{As $(C X,X \times \{ 0 \})$ is a good pair.}) \\ & \cong {\tilde{H}_{n + 1}}(C X,X \times \{ 0 \}) \quad (\text{As $X \neq \varnothing$.}) \\ & \cong {\tilde{H}_{n}}(X \times \{ 0 \}) \quad (\text{As explained above.}) \\ & \cong {\tilde{H}_{n}}(X). \end{align*} The claim is therefore established.

The definition of a good pair can be found on Page 114 of Allen Hatcher’s Algebraic Topology. On Page 118 of the same book, there is an explanation of the fact that $$\forall n \in \mathbb{Z}: \quad {\tilde{H}_{n}}(X,A) \cong {H_{n}}(X,A)$$ for all pairs $(X,A)$ such that $A \neq \varnothing$.

More generally let $(X, A)$ be a pair and consider the triple

$$(0 \times I) \cup (I \times A) \subset (\partial I \times X) \cup (I \times A) \subset I \times X.$$

One has from the associated exact sequence a boundary homomorphism

$$\partial_n : h_n(I \times X, (\partial I \times X) \cup (I \times A)) \longrightarrow h_{n-1}((\partial I \times X) \cup (I \times A), (0 \times I) \cup (I \times A))$$

which is in fact an isomorphism: the inclusion $(0 \times X) \cup (I \times A) \hookrightarrow I \times X$ has a retract $I \times X \to (0 \times X) \cup (I \times A)$ given by $(t, x) \mapsto (0, x)$; it follows that $h_*(I \times X, (0 \times X) \cup (I \times A)) = 0$ and looking at the long exact sequence shows the claim.

Further, there is a canonical isomorphism $h_*((\partial I \times X) \cup (I \times A), (0 \times X) \cup (I \times A)) \stackrel{\sim}{\to} h_*(X, A)$. In fact by excision on the subset $0 \times X \subset (0 \times X) \cup (I \times A))$, one has an isomorphism

$$\alpha : h_*((1 \times X) \cup (I \times A), I \times A) \stackrel{\sim}{\longrightarrow} h_*((\partial I \times X) \cup (I \times A), (0 \times X) \cup (I \times A))$$

and further the inclusion $(1 \times X, 1 \times A) \hookrightarrow ((1 \times X) \cup (I \times A), I \times A)$ has a retract defined by the restriction of $(t, x) \mapsto (1, x)$, and hence induces an isomorphism

$$\beta : h_*(X, A) \stackrel{\sim}{\longrightarrow} h_*(1 \times X, 1 \times A) \stackrel{\sim}{\longrightarrow} h_*((1 \times X) \cup (I \times A), I \times A).$$

Finally composing $\beta$, $\alpha$, and $\partial_{n+1}^{-1}$ gives an isomorphism

$$\sigma : h_n(X, A) \stackrel{\sim}{\longrightarrow} h_{n+1}(I \times X, (\partial I \times X) \cup (I \times A)).$$

In the case $A$ is a point $x_0 \in X$ and $(X, x_0)$ is well-pointed, one gets the desired isomorphism

$$\tilde{h}_n(X) \stackrel{\sim}{\longrightarrow} h_{n+1}(I \times X, (\partial I \times X) \cup (I \times x_0)) \stackrel{\sim}{\longrightarrow} \tilde{h}_{n+1}((I \times X)/((\partial I \times X) \cup (I \times x_0)).$$

• It doesn't seem like you interpreted the problem according to my edit. I meant $\Sigma X$ as wikipedia's $S X$. – user71500 Apr 8 '13 at 1:51
• Ah, I don't know how to do it without Mayer-Vietoris then. Since it follows quite easily from M-V, and since M-V is derived pretty easily from the axioms, maybe it's worth learning M-V first? – user314 Apr 8 '13 at 7:55