I have been studying the mod 2 Steenrod algebra. And I try to solve some exercises of it.
Can you help me to check this proof:
Let $SX$ denote the suspension of $X$, and let $S: \underline{H}^q(X) \rightarrow \underline{H}^{q+1}(SX)$ denote the suspension isomorphism. Then, from the property:
If $\delta: H^q(A) \rightarrow H^{q+1}(X,A)$ is the coboundary map, then $\delta \operatorname{Sq}^i = \operatorname{Sq}^i \delta$
Prove that $s \operatorname{Sq}^i = \operatorname{Sq}^i s$.
Proof: Let $CX$ and $C'X$ be two cones on $X$. Then $SX = CX \cup C'X$, where $CX \cap C'X =X$. The suspension isomorphism is defined by the following commutative diagram of reduced cohomology groups
$$\require{AMScd} \begin{CD} \underline{H}^q(X) @>s>> \underline{H}^{q+1}(SX) \\ @V{\cong}VV & @A{\cong}AA \\ \underline{H}^{q+1}(CX,X) @>{excisio}>> \underline{H}^{q+1}(SX,C'X) \end{CD}$$
Then, $s: \underline{H}^q(X) \rightarrow \underline{H}^{q+1}(CX,X) \cong \underline{H}^{q+1}(SX)$, It is easy to check that $s$ is a coboundary map. So by applying to the property, the statement is proved.
I am not sure about the last argument. Can you help me to check that?