Reading about Steenrod squares and a result regarding the Hopf invariant the following homeomorphism is used in a proof without being proved $\Sigma^k(C_f) = C_{\Sigma^kf}.$ Here $\Sigma$ is the reduced suspension and $C_f$ the mapping cone. I have written down a "proof" of this but it is hand-wavy to say the least.
So, assume $k=1$ (rest follows by induction). I argue that $\Sigma C X = C\Sigma X$, where $CX$ is the reduced cone, i.e. $CX = X\times I / \{x_0\}\times I\cup X \times\{1\}$.
By writing out the definitions I find $C\Sigma X = X\times I \times I \big/ \{x_0\}\times I\times I \cup X\times \{0\}\times I\cup X\times\{1\}\times I\cup X\times I \times \{1\}$ and $\Sigma C X = X\times I \times I \big/ \{x_0\}\times I\times I\cup X\times I\times\{0\}\cup X\times I\times\{1\}\cup X\times\{1\}\times I.$ Then $[x, t_1, t_2] \rightarrow [x, t_2, t_1]$ should be a homemorphism.
Continuing with $\Sigma C_f$ and writing out relations: $\Sigma C_f = \Sigma Y \amalg \Sigma C X \big/ \left([x,0,t]\sim[f(x), t]\right) = C_{\Sigma f}$ where the last equality follows from the homeomorphism above. Is this OK?