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Browder's book "Surgery on simply-connected manifolds" defines the Kervaire invariant in a very general setting. My question is: how does one get the more usual definition of the invariant for a framed (or Wu-orinented) manifold from Browder's general definition?

Browder's setting: For Poincare pairs $(X,Y)$ and $(A,B)$ with specified Spivak normal bundles, a normal map is a map $f: (X,Y) \to (A,B)$ that has degree 1 and is covered by a bundle map $b$ from one Spivak normal bundle to the other. Browder manages to define the $\mathbb Z /2$-valued Kervaire invariant $\sigma$ for any such map (no additional information needed), at least as long as $f^*$ takes the Wu class of $A$ to the Wu class of $X$. I find it hard to understand his definition.

Some simplification: For the sake of understanding, it should be safe to replace all the Poincare pairs with closed manifolds, and Spivak normal bundles with normal bundles. Then, a normal map becomes a degree-1 map of degree 1 between manifolds embedded in a high-dimensional Euclidean space that is covered by a bundle isomorphism of normal bundles. If the manifolds can be framed (or Wu-oriented), the Wu classes will be zero, so there's no reason to worry about them.

Question: How does this situation relate to the more usual situation of having a manifold with a framing or Wu orientation (which are, of course, necessary to define the Kervaire invariant in the usual way)? I'd imagine that there is some standard thing I could fix $(A,B)$ to be so that Browder's version of Kervaire invariant computes the usual Kervaire invariant of $(X,Y)$. I couldn't figure out what it should be. This choice of $(A,B)$ should somehow encode the Wu orientation on $(X,Y)$.

Please tell me if there are any mistakes in the above. Thank you!

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# Definition of the Kervaire invariant for normal maps (as in Browder's book)

Browder's book "Surgery on simply-connected manifolds" defines the Kervaire invariant in a very general setting. My question is: how does one get the more usual definition of the invariant for a framed (or Wu-orinented) manifold from Browder's general definition?

Browder's setting: For Poincare pairs $(X,Y)$ and $(A,B)$ with specified Spivak normal bundles, a normal map is a map $f: (X,Y) \to (A,B)$ that has degree 1 and is covered by a bundle map $b$ from one Spivak normal bundle to the other. Browder manages to define the $\mathbb Z /2$-valued Kervaire invariant $\sigma$ for any such map (no additional information needed), at least as long as $f^*$ takes the Wu class of $A$ to the Wu class of $X$. I find it hard to understand his definition.

Some simplification: For the sake of understanding, it should be safe to replace all the Poincare pairs with manifolds, and Spivak normal bundles with normal bundles. Then, a normal map becomes a map of degree 1 between manifolds embedded in a high-dimensional Euclidean space covered by a bundle isomorphism of normal bundles. If the manifolds can be framed (or Wu-oriented), the Wu classes will be zero, so there's no reason to worry about them.

Question: How does this situation relate to the more usual situation of having a manifold with a framing or Wu orientation (which are, of course, necessary to define the Kervaire invariant in the usual way)? I'd imagine that there is some standard thing I could fix $(A,B)$ to be so that Browder's version of Kervaire invariant computes the usual Kervaire invariant of $(X,Y)$. I couldn't figure out what it should be. This choice of $(A,B)$ should somehow encode the Wu orientation on $(X,Y)$.

Please tell me if there are any mistakes in the above. Thank you!