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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?

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I think once you have established $\Sigma C(X) \cong C(\Sigma X)$ there is a more general argument you can follow to obtain the result. This goes as follows:

A mapping cone of a morphism is obtained by forming the pushout diagram:

$$\begin{array} =X & \stackrel{f}{\longrightarrow} & Y \\ \downarrow && \downarrow \\ C(X) & \longrightarrow & C(f) \end{array} $$

Then since the functor $\Sigma$ is left adjoint to the loop space functor it follows that $\Sigma$ commutes with pushouts, i.e., applying $\Sigma$ to the above diagram again yields a pushout diagram

$$\begin{array} 0\Sigma X & \stackrel{\Sigma f}{\longrightarrow} & \Sigma Y \\ \downarrow && \downarrow \\ \Sigma C(X) & \longrightarrow & \Sigma C(f) \end{array} $$

Now using the homeomorphism $\Sigma C(X) \cong C(\Sigma X)$ you can see that $\Sigma C(f)$ is a model for the Cone of the morphism $\Sigma(f)$.

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Do you have a reference to prove $\Sigma C(X)\simeq C(\Sigma X)$? –  Sigur Oct 12 '12 at 0:36

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