# Do Steenrod Squares have naturality with homomorphisms that don't come from continous maps

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.

• 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 $\mathbb{R}P^\infty$. Jun 26 at 13:36
• 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 $H^*(\mathbb{R} P^n;\mathbb{Z}_2)$. Jun 26 at 14:53
• 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. Jun 26 at 14:55

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 $$n$$.