Suppose that X is a cell complex with $\tilde{H}_{*}(X) = 0.$ Prove that the suspension $SX$ is contractible. The question is:
Suppose that X is a cell complex with $\tilde{H}_{*}(X) = 0.$ Prove that the suspension $SX$ is contractible.
I feel like this link contains a part (or maybe all) of the solution to this question, am I correct? if so I just need a recap of the general idea of the solution please. If not could you give me a hint for the solution? 
Suspension: if $X$ is $(n-1)$-connected CW, is $SX$ $n$-connected?
 A: Sketch of solution :
1) the hypothesis implies that $X$ is connected (i.e. $0$-connected)
2)The given link then implies (or just by Van Kampen) $SX$ is $1$-connected
3) The suspension isomorphism, the Hurewicz theorem and the Whitehead theorem allow us to conclude 
A: Consider the map $f: \{*\} \to SX$ sending the single point to one of the 0-cells of $X$, which lies in $SX$ (So $f$ is WLOG an inclusion of a zero-cell). Then consider $f_*: \tilde{H}_n(\{*\}) \to \tilde{H}_n (SX)$. The first of these groups is zero for all $n$, and the second of these satisfies $\tilde{H}_n (SX) \cong \tilde{H}_{n-1}(X) \cong 0$ for all $n\geq 1$. For $n=0$, by hypothesis we know that $X$ is path-connected (because $\tilde{H}_0(X) \cong 0$), so $SX$ is too, and thus $\tilde{H}_0(SX) \cong 0$ as well. Since $f_*$ is a map of trivial groups it's automatically an isomorphism.
So we know that $f_*$ is an isomorphism on all reduced homology groups. It's also an isomorphism on all (absolute) homology groups. (To see why, note that we only need to check the case $n=0$ because reduced homology coincides with the absolute homology in positive degrees. But in degree zero, the induced map on absolute homology is given by $f_* \oplus \text{id}_\mathbb{Z}: \tilde{H}_0(\{*\}) \oplus \mathbb{Z} \to \tilde{H}_0(SX)\oplus \mathbb{Z}$ which is the identity map on $\mathbb{Z}$ since the reduced homology groups are zero.)
Now we know $f_*$ induces isomorphisms on all homology groups. Then since $X$ is path-connected, we have that $SX$ is simply connected (this is in the question you linked). Certainly the one-point space is also simply connected. So $\pi_1(SX, \{*\}) \cong 0$. Then the Hurewicz theorem tells you that the first nonzero relative homotopy and homology groups of the pair $(SX, \{*\})$ coincide. But all of the relative homology groups are zero, so $\pi_n(SX, \{*\}) \cong H_n(SX, \{*\}) \cong \tilde{H}_n(SX) \cong 0$ for all $n$.
Then since $f$ is an inclusion, we can look at the long exact homotopy sequence for the pair to see that $f$ induces an isomorphism on all homotopy groups (remember that the map induced by inclusion is one of the maps in the long exact sequence). Then Whitehead's theorem asserts that $f$ is a homotopy equivalence.
