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Recall that the surgery structure set of a smooth, compact n-manifold $M$ with normal bundle $\eta$ is the set of maps $f:N \rightarrow M$ that are homotopy equivalences up to the relation $f:N \rightarrow M $ is equivalent to $f':N' \rightarrow M$ if there is an h-cobordism $(W,N,N')$ with a map $(W,N,N') \rightarrow (M \times I, M \times \{0\}, M \times \{1\})$, restricting to the maps involved. This definition is from Ranicki's "Algebraic and Geometric Surgery".

We denote the structure set $\mathscr{S}(M)$.

This set of equivalence classes is of interest because it sits in the exact sequence of pointed sets $\mathscr{S}(M) \rightarrow \mathscr{N}(M) \rightarrow L_n(\pi_1(M))$, where $\mathscr{N}(M)$ is the set of normal invariants of $M$, i.e. the degree 1 maps from manifolds with lifts to fiberwise isomorphisms of the normal bundles, up to cobordisms which extend the map on normal bundles, so-called normal cobordisms.

The $L$ groups are the surgery obstruction groups, but these are not relevant to my question. My question is about how we can define the map $\mathscr{S}(M) \rightarrow \mathscr{N}(M)$ with no normal bundle information.

It turns out that between compact manifolds, a homotopy equivalence can always be covered by fiberwise isomorphism of normal bundles. Hence, we can always take our homotopy equivalence and make a $choice$ to turn it into a normal map. However, these choices are in general not fiberwise homotopic (e.g. $S^1$ has multiple different framings of the normal bundle).

So for this map to be well defined, it needs to be the case that we can find a normal cobordism between any two different lifts of the homotopy equivalence to the normal bundle. Why is this true?

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    $\begingroup$ Your definition of $N(M)$ is off: a class in $N(X)$ for a Poincaré complex $X$ is represented by a degree 1 map $N\rightarrow X$ from a manifold covered by a bundle map $TN\oplus \varepsilon^k\rightarrow \xi$ for some $k\ge0$ and \emph{some} bundle $\xi$ over $X$. If $X$ happens to be a manifold itself, $\xi$ need not agree with $TX$. $\endgroup$
    – archipelago
    Commented Sep 10, 2020 at 18:14
  • $\begingroup$ @archipelago Ah, so then is the map just if we have a homotopy equivalence we can push the normal bundle of the domain manifold forward and have a canonical normal map? $\endgroup$
    – Connor Malin
    Commented Sep 10, 2020 at 18:27
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    $\begingroup$ You pull back the tangent bundle of $N$ along a homotopy inverse. All this can be reformulated in terms of normal bundles, but I prefer the tangential perspective. $\endgroup$
    – archipelago
    Commented Sep 10, 2020 at 18:30
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    $\begingroup$ I do not quite follow your terminology but the claim "homotopy equivalence can always be covered by fiberwise isomorphism of [stable] normal bundle" in incorrect. Do a web search "non-tangential homotopy equivalences". What is true is that homotopy equivalence (of closed manifolds) pulls back the stable spherical fibrations. $\endgroup$
    – Igor Belegradek
    Commented Sep 10, 2020 at 19:43
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    $\begingroup$ Ranicki's book is nice but I would start studying surgery with the original papers by Milnor-Kervaire, Browder, and Novikov, who take the time to explain things. In particular, Browder's textbook on simply-connected surgery is a good start. $\endgroup$
    – Igor Belegradek
    Commented Sep 10, 2020 at 20:39

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