Does a suspension have the same homology as a reduced suspension?

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?

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 $$\bigoplus\limits_{i \in \mathbb{Z}} \mathbb{Z}$$.
• For your first paragraph : more generally it is true if $X$ is well-pointed, because then $I\to S X$ is still a cofibration Jun 13 '19 at 11:34