I was reading about the Steenrod Squaring operations in Milnor and Stasheff's, Characteristic Classes, now there is an axiom regarding naturality that says that given a continous map $f:(X,Y)\rightarrow (X',Y')$ and $f^*$ is the induced map on cohomology groups then $Sq^i\circ f^*=f^*\circ Sq^i$.
Given a map $g^*$ between cohomology groups that does not come from a continous map between topological spaces do we still get the naturality with the Steenrod Squares ?
The motivation behind this is that, assuming coefficients in $\mathbb{Z}_2$, there is a monomorphism from the cohomology of the real grassmannian to the cohomology of $\mathbb{R}P^\infty\times ...\times\mathbb{R} P^\infty$, as the grassmannian is a universal bundle and the squaring operations are easily computable on $\mathbb{R}P^\infty$ this would allow a description for the action of the Steenrod Squares on any real vector bundle. Unfortunately, I am not sure there is a continous map that induces said monomorphism.

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    $\begingroup$ The monomorphism you describe is just induced by the map $\mathbb{R}P^\infty\times\dots\times\mathbb{R}P^\infty$ to the Grassmannian that classifies the direct sum of the canonical line bundles from each $\mathbb{R}P^\infty$. $\endgroup$ Jun 26 at 13:36
  • $\begingroup$ actually the monomorphism I was thinking about was one that arises in the computation of the cohomology of the grassmannian, it takes a monomial of Stiefel-Whitney classes of tautological bundle over the grassmannian, i.e. $\mathbb{Z}_2[w_1(\gamma^n),...,w_n(\gamma^n)]$, to the symmetric polynomials of the polynomial algebra $\mathbb{Z}_2[a_1,...,a_n]$ where $a_i$ generate each $H^*(\mathbb{R} P^n;\mathbb{Z}_2)$. $\endgroup$
    – AOJIDSOeoi
    Jun 26 at 14:53
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    $\begingroup$ That is the same map, since the Stiefel-Whitney classes of a direct sum of line bundles are the elementary symmetric polynomials in the first Stiefel-Whitney classes of the line bundles. $\endgroup$ Jun 26 at 14:55

1 Answer 1


No, this is certainly not true. For instance, take any space $X$ for which $Sq^i:H^n(X)\to H^{n+i}(X)$ is nontrivial for some $i$ and $n$, and consider $g^*:H^*(X)\to H^*(X)$ which is $0$ in degree $n$ but the identity in degree $n+i$. Then $g^*\circ Sq^i$ is nonzero in degree $n$ but $Sq^i\circ g^*$ is $0$ in degree $n$.


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