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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?

Eric Wofsey
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user69833
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1 Answers1

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That's not quite right. $s$ itself isn't directly a coboundary map. However you know that cohomology operations are natural maps, so they also commute with maps of the form $f^* : H^i(X,A) \to H^i(Y,B)$ where $f : (Y,B) \to (X,A)$ is a map of pairs. So $\operatorname{Sq}^i$ commutes with:

  • the coboundary map $H^q(X) \to H^{q+1}(CX,X)$;
  • $H^{q+1}(SX,C'X) \to H^{q+1}(CX, X)$ induced by the inclusion $(CX,X) \subset (SX, C'X)$; by excision this is an isomorphism so you can invert it, and if $\operatorname{Sq}^i f = f \operatorname{Sq}^i$ then $f^{-1} \operatorname{Sq}^i = \operatorname{Sq}^i f^{-1}$;
  • $H^{q+1}(SX,C'X) \to H^{q+1}(SX)$ induced by the inclusion $(SX, \varnothing) \subset (SX, C'X)$;

so $\operatorname{Sq}^i$ commutes with the composite of all three maps, that is $s : H^q(X) \to H^{q+1}(SX)$.

Najib Idrissi
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