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it is a type, you are right, the point is that it is in fact a morphism of gradd abelian groups, so if one forgets that cohomology is a ring then it is a morphism, but since it is just a collection of natural transformations, which may not be how you want to think about it pedagogically, it is really
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The
Steenrod squares are just some subgroup of this. (If I remember correctly, they aren't the entire cohomology of E-M space, but a particularly easy subgroup to calculate. Note we restrict to dimensions where n \leq m.) The squares themselves are a basis for this subgroup, and the Adem relations
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can be calculated on E-M space, it follows they hold for all manifolds. Maybe someone else knows about this and can sat a little more, especially about why you pick this particular subgroup.
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The coaction is jot induced by the product on E but by the unit from S to E. The latter would induce an action since the second variable of hom is covariant.
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H^n+i(X). and i will fix this now, but what was meant was that it is not a morphism of rings, it is almost never a ring homomorphism, see the cartan fmla.
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so I deduce that cohomology operations need to be morphisms in the category of groups, i.e. group homomorphisms.
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Relate to computing the cohomology of all mod p eilenberg-maclane spaces (given in
Hatcher spectral sequences)
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Discuss some of the computations of Adams spectral sequence coming from
Steenrod squares (McCleary)
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on
Knowledge. If you would like to participate, please visit the project page, where you can join
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This page should discuss examples of the steenrod squaring operation. This should inlcude
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Then the
Steenrod squares are induced by composing with homotopy classes of maps
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I never really understood
Steenrod squares until somebody told me this. Let
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155:"...Note that cohomology operations need not be group homomorphisms."
449:{\displaystyle K(\mathbb {Z} _{2},m)\to K(\mathbb {Z} _{2},m+n)}
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only a morphism from the abelian group H^n(X) ---: -->
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518:{\displaystyle H^{n+m}(K(\mathbb {Z} _{2},n)).}
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237:{\displaystyle K(\mathbb {Z} _{2},m)}
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369:{\displaystyle H^{m}(X)\cong .}
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