4
$\begingroup$

If $X$ is a space, the suspension $SX$ is defined to be the image of $X \times [0, 1]$ under the quotient identifying each of $X \times \{0\}$ and $X \times \{1\}$ to single points. The reduced suspension $\Sigma X$ is then obtained by taking the image of $SX$ under the quotient identifying some line $L = x_0 \times [0, 1]$ to a single point. Does it follow that $SX$ and $\Sigma X$ have the same reduced homology groups for $n \geq 0$?

If $(\Sigma X, L)$ is a good pair, then we can apply the long exact sequence of relative homology groups, and observing that $\tilde H_n(L) = 0$ for all $n$, we get isomorphisms $\tilde H_n(SX) \simeq \tilde H_n(SX / L)$ for all $n$, and $SX/L = \Sigma X$. Is this reasoning correct?

$\endgroup$
4
$\begingroup$

If the space is a CW complex this is true since you can supply a CW structure so that the interval you collapse is a subcomplex, and so the two are homotopy equivalent.

There is a counterexample for general spaces. Let $X$ be the space $\{1/n| n \in \mathbb{N}\}\cup \{0\}$. Then $\Sigma X$ is the Hawaiian earring which has homology that is not free. $SX$ has much simpler homology since you can use the Mayer-Vietoris sequence to show its homology is just the homology of $X$ shifted up, so it is $\bigoplus\limits_{i \in \mathbb{Z}} \mathbb{Z}$.

$\endgroup$
  • 1
    $\begingroup$ For your first paragraph : more generally it is true if $X$ is well-pointed, because then $I\to S X$ is still a cofibration $\endgroup$ – Max Jun 13 at 11:34

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.