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.

| cite | improve this answer | |

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 $.

| cite | improve this answer | |

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

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

| cite | improve this answer | |
  • $\begingroup$ It doesn't seem like you interpreted the problem according to my edit. I meant $\Sigma X$ as wikipedia's $S X$. $\endgroup$ – user71500 Apr 8 '13 at 1:51
  • $\begingroup$ 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? $\endgroup$ – user314 Apr 8 '13 at 7:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.