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The exercise is the following:

Show that, for any homology theory (satisfying the usual axioms), there is a natural isomorphism $ \tilde{H_i}(X) \rightarrow \tilde{H}_{i+1}(\Sigma X)$.

Well, I tried using the long exact sequence:

$$...\rightarrow \tilde{H}_{i+1}(X_{1/2}) \rightarrow \tilde{H}_{i+1}(\Sigma X) \rightarrow H_{i+1}(\Sigma X, X_{1/2}) \rightarrow \tilde{H}_{i}(X_{1/2}) \rightarrow \tilde{H}_{i}(\Sigma X) \rightarrow ...$$

where $X_{1/2}$ is $\{1/2\} \times X$ in the suspension. Then, I tried to compute $H_{i+1}(\Sigma X, X_{1/2})$. For that, I enlarged a bit $X_{1/2}$ to $\overline{X}_{1/2}:=[\frac{1}{4}, \frac{3}{4}] \times X$ and by excision (cutting off a small neighbourhood of $X_{1/2}$) and the long exact sequence for the pair $(C, X_{3/4})$, where $C$ is the upper part of the "cone" that is left, I managed to prove that:

$$H_{i+1}(\Sigma X, X_{1/2}) \cong H_{i}(X) \oplus H_i(X)$$

but I got stuck after this.

Aloizio Macedo
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4 Answers4

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I'm going to use a different notation mostly because I'm largely copying this out of an old homework of mine.

Let $C_+^n$ and $C_-^n$ be the cones in $\Sigma X$, let $U$ be a neighborhood around the cone point with $\overline{U}\subset Int(C_-^n)$. Then $U$ can be excised, so

$$\widetilde{H}_q(\Sigma X,C_-^n)\simeq \widetilde{H}_q(\Sigma X\setminus U,C_-^n\setminus U).$$

Since $(\Sigma X\setminus U,C_-^n\setminus U)$ deformation retracts to $(C_+^n,X)$ we get

$$\widetilde{H}_q(\Sigma X,C_-^n)\simeq\widetilde{H}_q(C_+^n,X)$$

Since $CX$ is contractible, $C_+^n\simeq C_-^n\simeq \{pt.\}$, and thus $\widetilde{H}_\ast(C_+^n)\simeq\widetilde{H}_\ast(C_-^n)\simeq 0$. Then the long exact sequence

$$ \dots\to \widetilde{H}_q(C_-^n)\to\widetilde{H}_q(\Sigma X)\to \widetilde{H}_q(\Sigma X,C_-^n)\to\widetilde{H}_{q-1}(C_-^n)\to\dots $$

gives

$$ \dots\to 0 \to\widetilde{H}_q(\Sigma X)\to \widetilde{H}_q(\Sigma X,C_-^n)\to 0 \to\dots $$

so that $\widetilde{H}_q(\Sigma X)\simeq\widetilde{H}_q(\Sigma X, C_-^n)$. Similarly, the long exact sequence

$$ \dots\to \widetilde{H}_q(C_+^n)\to\widetilde{H}_q(C_+^n,X)\to \widetilde{H}_{q-1}(X)\to\widetilde{H}_{q-1}(C_+^n)\to\dots $$

gives

$$ \dots\to 0 \to\widetilde{H}_q(C_+^n,X)\to \widetilde{H}_{q-1}(X)\to 0 \to\dots $$

so that $ \widetilde{H}_q(C_+^n,X)\simeq \widetilde{H}_{q-1}(X) $.

Thus, $\widetilde{H}_q(\Sigma X)\simeq\widetilde{H}_q(\Sigma X, C_-^n)\simeq\widetilde{H}_q(C_+^n,X)\simeq \widetilde{H}_{q-1}(X) $.

Joe Moeller
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    I know it has been a long time, but could you please explain why $(\Sigma X\setminus U,C_-^n\setminus U)$ deformation retracts to $(C_+^n,X)$? – John Doe May 14 '19 at 23:15
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    You should imagine $\Sigma X$ as a sphere, but where the equator looks like $X$. Then $C^n_+$ is the northern hemisphere, $C^n_-$ is the southern hemisphere, and $U$ is a small neighborhood of the south pole. Then you could imagine "simultaneously" deformation retracting $\Sigma X\setminus U$ to the northern hemisphere, and $C^n_-\setminus U$ to the equator. – Joe Moeller May 16 '19 at 23:00
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Let $CX$ be the cone on $X$. We can write $\Sigma X$ as the following pushout: $$\require{AMScd} \begin{CD} (x_0\times I)\cup (X\times\{1\}) @>>> * \\ @VVV @VVV \\ CX @>>> \Sigma X \end{CD}$$ where the left map is the inclusion. The reduced Mayer Vietoris sequence corresponding to that pushout contains the following segment:

$$\tilde{H}_{i+1}(*)\oplus \tilde{H}_{i+1}(CX) \to \tilde{H}_{i+1}(\Sigma X) \to \tilde{H}_{i}((x_0\times I)\cup (X\times\{1\})) \to \tilde{H}_{i}(*)\oplus \tilde{H}_{i}(CX)$$

Note that $(x_0\times I)\cup (X\times\{1\})$ deformation retracts onto $X$, and $CX$ is contractible. Then the above sequence reads

$$0\to \tilde{H}_{i+1}(\Sigma X) \to \tilde{H}_{i}(X)\to 0.$$

This gives the desired isomorphism.

iwriteonbananas
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Let $\Sigma X = X \times [0,1]$, with $X \times \{0\}$ identified to a point and $X \times \{1\}$ identified to a point. Let $A = X \times [0,\frac{3}{4}], B = X \times [\frac{1}{4},1]$ with the same identifications. Then $A$ and $B$ are contractible, and $A \cap B \simeq X$. Reduced Mayer-Vietoris gives the long exact sequence $$\cdots \rightarrow \tilde H_{i+1}(A) \oplus \tilde H_{i+1}(B) \rightarrow \tilde H_{i+1}(\Sigma X) \rightarrow \tilde H_i(A \cap B) \rightarrow \tilde H_i(A) \oplus \tilde H_i(B) \rightarrow \cdots$$ and for all $i$, this is $$\cdots \rightarrow 0 \rightarrow \tilde H_{i+1}(\Sigma X) \rightarrow \tilde H_i(X) \rightarrow 0 \rightarrow \cdots$$ hence giving the desired isomorphism.

Jonathan Michala
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Let $Y = CX = X \times I /X \times \{1\}$ and $A = X\times \{0\}$ a closed subspace of $Y$. By this definition it follows that $Y/A = \Sigma X$.

The homology groups are as follows: $\widetilde{H}_n(A) = \widetilde{H}_n(X)$ as the spaces $A$ and $X$ are homotopic and $\widetilde{H}_n(Y) = 0$ as the cone of a space is contractible.

Thus by Hatcher 2.13 the following sequence is exact $$ \cdots \rightarrow \widetilde{H}_n(Y) \rightarrow \widetilde{H}_n(Y/A) \rightarrow \widetilde{H}_{n-1}(A) \rightarrow \widetilde{H}_{n-1}(Y) \cdots $$ and as $\widetilde{H}_n(Y) = 0$ the parts of the long exact sequence being $$ 0 \rightarrow \widetilde{H}_n(Y/A) \rightarrow \widetilde{H}_{n-1}(A) \rightarrow 0 $$ imply an isomorphism between the groups $\widetilde{H}_n(Y/A) \cong\widetilde{H}_{n-1}(A)$ which as stated above are the same as $\widetilde{H}_n(\Sigma X) \cong\widetilde{H}_{n-1}(X)$.

Jakob
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