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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?

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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 $\bigoplus\limits_{i \in \mathbb{Z}} \mathbb{Z}$.

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    $\begingroup$ For your first paragraph : more generally it is true if $X$ is well-pointed, because then $I\to S X$ is still a cofibration $\endgroup$ Commented Jun 13, 2019 at 11:34

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