Let $X$ be a simply connected space of dimension $4k, k \geq 2$ for which the Poincaré duality theorem holds. In the previous post, I stated a theorem of Browder and Novikov which provides necessary and sufficient conditions for $X$ to be homotopy equivalent to a smooth manifold.

• $X$ must admit a candidate $\xi$ for a stable normal bundle (a lift of the Spivak normal fibration).
• Hirzebruch’s signature theorem, with this proto-normal bundle fed in, should accurately compute the signature of $X$.

The goal of this post is to sketch a proof of this theorem. The proof proceeds in two steps. The first step is to produce a degree one normal map $f: M \to X$ from a smooth manifold, i.e. a map which pulls $\xi$ back to the stable normal bundle of $M$ and which preserves the fundamental class. The second (and harder) step is to do surgery on $f$, to make it a homotopy equivalence. This surgery can be done “formally” before the middle dimension, but at the middle dimension a more careful bookkeeping process is required. It is here that the signature theorem, together with facts about quadratic forms over the integer, become necessary.

These are essentially the second half of the notes I prepared for the Kan seminar, for my second talk. My second talk was officially on Browder’s paper “Homotopy type of differentiable manifolds,” which announces (without proof) this result. In practice, I found the Kervaire-Milnor paper “Groups of homotopy spheres I” and the Milnor paper “A procedure for killing the homotopy groups of differentiable manifolds” very helpful in learning about this material.

1. Degree one normal maps

Let ${X}$ be a Poincaré complex satisfying the first condition of the statement of the Browder-Novikov theorem. Our goal is to produce a degree one normal map ${f: M \rightarrow X}$. In order to do this, we can use the Pontryagin-Thom construction. Namely, recall that we have a map

$\displaystyle \phi: S^N \rightarrow X^{\xi}$

inducing an isomorphism on top homology. Here ${X \hookrightarrow X^{\xi}}$ via the zero section, and a small neighborhood of ${X}$ in ${X^{\xi}}$.

Using a transversality argument, we can make ${\phi}$ transverse to the zero section. In this case, ${M \stackrel{\mathrm{def}}{=}\phi^{-1}(X)}$ is a manifold, whose normal bundle (in ${S^N)}$ is isomorphic to the pull-back ${\xi}$, under the map ${f: M \rightarrow X}$ that we get. Moreover, we have a map of pairs

$\displaystyle (S^N, S^N \setminus M ) \rightarrow (X^{\xi}, X^{\xi} \setminus X)$

and a look at this map, and the naturality of the Thom isomorphism, shows that the map

$\displaystyle f: M \rightarrow X$

is a degree one normal map. That is, ${f_*([M]) = [X])}$; that follows by looking at the image of ${[S^N]}$ in ${H_*(X^{\xi}; \mathbb{Z})}$.

We find:

Proposition 1 If ${X}$ is a Poincaré complex, a lift ${\xi}$ of the Spivak normal fibration (equivalently, a bundle ${\xi}$ such that ${M^{\xi}}$ is ${S}$-reducible) produces a degree one normal map ${f: M \rightarrow X}$ from a manifold.

The Spivak normal fibration is a stable spherical fibration, and is therefore classified by a map

$\displaystyle X \rightarrow \varinjlim B\mathrm{Haut}(S^n, S^n),$

to the colimit of the classifying spaces of the homotopy self-equivalences of ${S^n}$. A choice of vector bundle lifting the Spivak fibration is a lift of the map ${X \rightarrow \varinjlim B\mathrm{Haut}(S^n, S^n)}$ under the J-homomorphism ${BSO \rightarrow\varinjlim B\mathrm{Haut}(S^n, S^n) }$.

2. Surgery below the middle dimension

Let ${X}$ be a Poincaré complex. Given a lift of the Spivak normal fibration to a vector bundle on ${X}$, we saw in the previous section that we get a degree one normal map ${f: M \rightarrow X}$. Our goal is to modify ${f}$ so as to make it closer to a homotopy equivalence; in order to do so we need a method of modifying the homotopy type of a manifold. This method is surgery.

Definition 2 Observe that

$\displaystyle \partial( S^p \times D^{q+1} ) = \partial(D^{p+1} \times S^q) = S^p \times S^q.$

${p}$-surgery on the (oriented) ${n}$-manifold ${M}$ consists of taking an (oriented) imbedding ${S^p \times D^{q+1}}$ (where ${p+q+1 = n}$), cutting out the image of the interior, and pasting in a copy of ${D^{p+1} \times S^q}$ along the common boundary. We get a new manifold ${M' =( M - S^p \times \mathrm{Int}D^{q+1}) \cup_{S^p \times S^q} D^{p+1} \times S^q}$.

Given any imbedding ${S^p \times D^{q+1} \hookrightarrow M}$, we can perform a ${p}$-surgery. We can also go in the other direction: after a ${p}$-surgery on ${M}$, we can always perform a ${q}$-surgery to get back to ${M}$ (where ${p+q + 1 = \dim M}$ as before).

The next definition shows that the result is always oriented cobordant to ${M}$.

Definition 3 The trace of a ${p}$-surgery as above consists of ${M \times I \cup_{S^p \times D^{q+1}} D^{p+1} \times D^{q+1}}$. This is a manifold-with-boundary (obtained from ${M \times I}$ by “attaching a handle”) whose boundary is the disjoint union of ${M}$ and the new manifold ${M'}$. Note that, up to homotopy, ${W \simeq M \cup_{S^p} D^{p+1}}$.

In fact, any two manifolds are related by a surgery if and only if they are oriented cobordant. This is a consequence of Morse theory.

Example 1 Let ${M}$, ${M'}$ be two ${n}$-dimensional manifolds. We can perform a zero-surgery on ${ M \sqcup M'}$ (based on any imbedding ${S^0 \rightarrow M \sqcup M'}$ which takes one point to ${M}$ and another to ${M'}$). The result is the connected sum ${M \# M'}$.

Example 2 The sphere ${S^2}$ has the property that the complement of a band of the equator is diffeomorphic to ${S^0 \times D^2}$. A 0-surgery replaces this with a ${D^1 \times S^1}$ and results in a torus. More generally, a 0-surgery on a compact orientable surface increases the genus by one.

Our goal is now to use surgery to start with the map ${f: M \rightarrow X}$ and replace it by a highly connected map. The degree of connectivity of the map can be measured by the homotopy groups ${\pi_i(X, M)}$: recall that these consist of homotopy classes of diagrams

Given such a diagram, we have a class ${\alpha \in \pi_{i-1}(M)}$; this is the image under the boundary map. We would like to do surgery on ${\alpha}$ in such a way to get a manifold ${M'}$ mapping to ${X}$ (in such a way that it is still a normal degree one map) but so that ${\pi_*(X, M')}$ is smaller than ${\pi_*(X, M)}$, at least in the lowest dimension. This can be done up to the middle dimension, when an obstruction arises. For one thing, it will not always be possible to do the surgery, and even when it is, we have to take care that it does not complicate rather than simplify our manifold.

We will prove:

Proposition 4 Consider a degree one normal map ${f: M \rightarrow X}$ between a ${2m}$-dimensional manifold ${M}$ and a ${2m}$-dimensional Poincaré complex ${X}$ with bundle ${\xi}$. Then, by surgery on ${f}$, we can produce a degree one normal map ${\widetilde{f}: \widetilde{M} \rightarrow X}$ such that ${\pi_i(X, \widetilde{M}) = \pi_i(\widetilde{f}) = 0}$ for ${i \leq m}$.

Proof: As above, the strategy is to perform surgery on ${f}$. Suppose ${i \leq m}$ is minimal such that ${\pi_i(X, M) \neq 0}$ (it’s possible ${i = 0}$), and choose a class ${\alpha \in \pi_i(X, M)}$, represented by a diagram as above. We will construct a degree one normal map ${f': M' \rightarrow X}$ where ${M'}$ is obtained by an ${(i-1)}$-surgery on ${M}$ so as to kill ${\alpha}$.

In order to arrange that ${M'}$ also maps to ${X}$, we will extend the normal map ${M \rightarrow X}$ to a normal map ${W \rightarrow X }$, where ${W}$ is the trace of the surgery on ${M}$. The benefit is that ${W}$ is obtained by attaching a cell from ${M \times I}$, so one knows when and how to extend maps from ${M \times I}$ to ${W}$: it is in particular a homotopy problem. The following definition will be useful:

Definition 5 bordism between degree one normal maps ${f: M \rightarrow X, f': M' \rightarrow X}$, from ${n}$-dimensional manifolds ${M, M'}$ to an ${n}$-dimensional Poincaré complex ${X}$ with a lift ${\xi}$ of the Spivak fibration, is a normal map ${F: (W, M, M') \rightarrow (X \times I, X \times \left\{0\right\}, X \times \left\{1\right\})}$, where ${W}$ is a manifold-with-boundary with ${\partial W = M \sqcup (-M')}$. In particular,

1. ${F}$ is required to extend ${f}$ and ${f'}$.
2. An identification of the stable normal bundle of ${W}$ with ${F^* \xi}$ is given, and it extends the identifications of ${f^* \xi, f'^* \xi}$.
3. ${F}$ carries the relative fundamental class in ${H_{n+1}( W, M \sqcup M'; \mathbb{Z})}$ into that of ${H_{n+1}( X \times I, X \times \partial I)}$.

The strategy is to start with ${M}$, and then to construct a bordism between the degree one normal map ${f: M \rightarrow X}$ and a better map ${f': M' \rightarrow X}$ (which will be obtained via surgery).

In order to do this, we will start by observing that ${\partial\alpha: S^{i-1} \rightarrow M}$ can be represented by an imbedding, in view of Whitney’s imbedding theorem. By abuse of notation, we will identify ${\alpha}$ with ${\partial \alpha}$, although ${\alpha}$ consists not only of ${\partial \alpha}$ but also of a trivialization of ${f \circ \partial \alpha}$.

We are given a choice of trivialization of ${f \circ \alpha}$, which determines a choice of trivialization of the composite

$\displaystyle S^{i-1} \stackrel{\alpha}{\rightarrow} M \stackrel{f}{\rightarrow} X \stackrel{\xi}{\rightarrow} BO,$

which classifies the stable normal bundle of ${\alpha}$ in ${M}$. We would like to choose a choice of trivialization of the unstable normal bundle of ${\alpha}$ in ${M}$ compatible with this.

Observe that since the stable normal bundle of ${\alpha}$ is trivial, so is the unstable one; in fact, the map

$\displaystyle \pi_{i-2}(SO(2m-i+1)) \rightarrow \pi_{i-2}(SO)$

is an isomorphism. Moreover, we can choose a trivialization of the normal bundle of ${\alpha}$ which agrees stably with the trivialization of ${(f \circ \alpha)^* \xi}$. We can do this because trivializations of the unstable normal bundle of ${\alpha}$ are a torsor over ${\pi_{i-1}(SO(2m-i+1))}$ while those of the stable bundle are a torsor over ${\pi_{i-1}( SO)}$, and

$\displaystyle \pi_{i-1}( SO(2m-i+1)) \rightarrow \pi_{i-1}(SO)$

is an isomorphism, as ${i \leq m}$.

This choice of trivialization of the unstable normal bundle of ${\alpha}$ determines an imbedding

$\displaystyle S^{i-1} \times D^{q+1} \hookrightarrow M, \quad q = \dim M - i.$

by the tubular neighborhood theorem. We then proceed to do surgery on this imbedding.

The resulting trace of the surgery is the manifold-with-boundary

$\displaystyle W = M \times I \cup_{S^{i-1} \times D^{q+1}} D^{i} \times D^{q+1}.$

We would like to claim two things about ${W}$:

1. The map ${M \times I {\rightarrow} X \times I}$ extends (canonically) to a map ${F: W \rightarrow X \times I}$ (carrying the boundary into ${X \times \left\{0, 1\right\}}$).
2. The identification of the stable normal bundle of ${M}$ with ${f^*\xi}$ extends to an identification of that of ${W}$ with ${F^*\xi}$.

In order to see (1), observe that ${W}$ has the homotopy type of an ${i}$-cell attached along ${\alpha}$. To extend ${M \rightarrow X}$ along ${W}$ is to give a null-homotopy of ${f \circ \alpha}$, which is precisely what we have been given.

In order to see (2), observe that the stable normal bundle on ${W}$ is specified by the bundle ${\nu \simeq f^*\xi}$ on ${M}$ and by a trivialization of ${\nu|_{S^{i-1}}}$: namely, by the trivialization given by the tubular neighborhood ${S^{i-1} \times D^{q+1}}$ of ${\alpha(S^{i-1})}$. But this trivialization coincides with the trivialization of ${f^*\xi}$ arising from the nullhomotopy of ${f \circ \alpha}$, by construction of the tubular neighborhood. Since these two trivializations coincide, we find that ${M' \rightarrow X}$ is still a normal map: even better, ${W \rightarrow X \times I}$ is a normal map. Moreover, the fundamental class in ${(W, \partial W)}$ maps to that of ${(X \times I, X \times \partial I)}$. This implies that the fundamental class of ${M' }$ goes to that of ${X}$.

What can we say about the map ${f': M' \rightarrow X}$ that we have produced? Observe that ${W}$ is obtained by attaching an ${i}$-cell along ${\alpha}$ and ${W \rightarrow X}$ is based on the nullhomotopy given by ${\alpha}$ itself. It follows that ${\alpha}$ is killed in ${\pi_i(X, W)}$. More precisely, the map

$\displaystyle \pi_k(X, M) \rightarrow \pi_k(X, W)$

is an isomorphism for ${k < i}$, and a surjection for ${k =i}$ with ${\alpha}$ in the kernel. Moreover, up to homotopy, ${W}$ is obtained from ${M'}$ by attaching a ${(q+1)}$-cell, where ${q = \dim M - i \geq i }$. It follows that ${\pi_k(X,M' ) \rightarrow \pi_k(X, W)}$ is an isomorphism for ${k \leq i}$. Putting these together, we find that

$\displaystyle \pi_k(X, M) \simeq \pi_k(X, M') , \quad k < i,$

while ${\pi_i(X, M')}$ is smaller than ${\pi_k(X, M)}$. Note moreover that ${M' \rightarrow X}$ preserves fundamental classes: we have in fact not modified the top homology group by passing to ${M'}$.

Continuing this process, we can kill all the homotopy groups below the “middle dimension.” $\Box$

Example 3 Let’s take ${X = S^{2m}}$. To give a degree one normal map ${f: M \rightarrow X}$ is equivalent to giving aframed ${2m}$-manifold (as the map ${f}$ comes from free by crushing the ${(2m-1)}$-skeleton). The claim is that one can perform framed surgery on ${M}$ so as to produce a ${(m-1)}$-connected manifold. In other words, every framed ${2m}$-manifold is framed cobordant to a framed ${(m-1)}$-connected manifold. If we could continue framed surgery and get an ${m}$-connected manifold, Poincaré duality would imply that we had a homotopy sphere. It turns out that, in all but finitely many dimensions, the obstruction to doing so vanishes, and every framed cobordism class is represented by a homotopy sphere. We will see that this happens for ${m}$ even in the next section. (For ${m}$ odd, the question is equivalent to the Kervaire invariant problem.)

3. The surgery obstruction

It follows that, given an ${n}$-dimensional (for ${n = 2m}$) simply connected Poincaré complex ${X}$ with a lift of the Spivak fibration to a vector bundle, we can produce a degree one normal map ${f: M \rightarrow X}$ which is ${m}$-connected. We now would like to find the obstruction to continuing this process. Note, however, that there is exactly one more step to getting a homotopy equivalence. This will follow from the next lemma and the following discussion.

Lemma 6 Let ${f: M \rightarrow X}$ be a degree one map of Poincaré complexes. Then for any abelian group ${G}$,$\displaystyle f_*: H_*(M; G) \rightarrow H_*(X; G)$

is a split surjection.

Proof: In fact, we observe that there is a commutative diagram

The splitting comes from chasing around the diagram, since the vertical maps are isomorphisms. $\Box$

Now suppose ${f: M \rightarrow X}$ is ${m}$-connected, for ${m = \frac{n}{2}}$. Then the Hurewicz theorem implies that ${H_i(X; M; \mathbb{Z}) = 0}$ for ${i \leq m}$ and that the Hurewicz map

$\displaystyle \pi_{m+1}(X, M) \rightarrow H_{m+1}(X , M; \mathbb{Z})$

is an isomorphism. In particular, we find that the split surjection ${H_i(M; \mathbb{Z}) \rightarrow H_i(X; \mathbb{Z})}$ is an isomorphism for ${i < m}$. If it were an isomorphism for ${i = m}$, then Poincaré duality would imply that we had a homotopy equivalence already.

So let ${K = K(f) = \ker( H_m(M; \mathbb{Z}) \rightarrow H_m(X; \mathbb{Z})) \simeq \pi_{m+1}(X, M)}$. The final step in the surgery process is to kill the group ${K}$. The theorem thus follows from:

Proposition 7 Surgery on ${f: M \rightarrow X}$ can kill ${K}$ if and only if ${X}$ satisfies Hirzebruch’s signature formula.

Proof: The necessity of ${X}$‘s satisfying the signature formula, in order to be homotopy equivalent to a manifold, is evident. Conversely, suppose that ${X}$ satisfies the signature formula. Then

$\displaystyle \sigma(M) = \sigma(X)$

since ${M \rightarrow X}$ is a normal map. We have a commutative diagram:

which identifies the group ${H^m(M; \mathbb{Z})}$ as the direct sum ${K \oplus H^m(X; \mathbb{Z})}$.

The claim is that ${K \perp H^m(X; \mathbb{Z})}$ under the bilinear form on ${H^m(M; \mathbb{Z})}$ given by the cup product. In fact, the form sends a pair ${(a,b) \in H^m(M; \mathbb{Z})}$ to the evaluation pairing

$\displaystyle \left \langle \phi_M^{-1}(a), b\right\rangle, \quad \phi_M^{-1}: H^m(M; \mathbb{Z}) \simeq H_m(M; \mathbb{Z}).$

If ${b = f^* b'}$ comes from ${b' \in H^m(X; \mathbb{Z})}$, then that’s equivalent to evaluating

$\displaystyle \left \langle \phi_M^{-1}(a), f^*b \right\rangle = \left \langle f_* \phi_M^{-1}(a) , b\right\rangle$

and consequently the elements of the form ${f^* b'}$ are orthogonal to the image of ${\phi_M(K) \subset H^m(X; \mathbb{Z})}$.

It follows that the cup product defines a nondegenerate symmetric bilinear form on ${K}$ itself. Its signature is zero, since ${\sigma(M) = \sigma(X)}$: we have decomposed ${H^m(X; \mathbb{Z}) = H^m(M; \mathbb{Z}) \oplus K}$ as groups with bilinear forms. The strategy is to use results from the theory of quadratic forms to show that ${K}$ can be killed by surgery.

Lemma 8 ${K}$ is a free abelian group.

Proof: For any abelian group ${G}$, we have ${H_i(X, M; G) = 0}$ for ${i < m+1}$, so ${H_i(M; G) \rightarrow H_i(X; G)}$ is an isomorphism for ${i < m}$. This implies that ${H^{n-i}(X; G) \rightarrow H^{n-i}(M; G)}$ is an isomorphism for such ${i}$, or ${H^r(X; G) \rightarrow H^r(M; G)}$ is an isomorphism for ${r > m}$. For example, taking ${r = m+1, m+2}$ we find that ${H^{m+2}(X, M; G) = 0}$. Using the universal coefficient theorem, it follows that there can be no torsion in ${K}$, or there would be an ${\mathrm{Ext}}$ term in ${H^{m+2}(X, M; G) = 0}$. $\Box$

It follows that ${K}$ is a free abelian group, on which there is a nonsingular bilinear pairing with signature zero. The rest of the proof proceeds by using facts about quadratic forms to write ${K}$ in a particularly convenient form. Then, a careful bookkeeping of the surgery process shows that we can kill (spherical representatives of the) the generators of ${K}$, one by one.

Let’s recall the following definition.

Definition 9 lattice is a free, finitely generated abelian group ${Q}$ together with a symmetric bilinear form ${Q \times Q \rightarrow \mathbb{Z}}$ which is unimodular (i.e., establishes an isomorphism ${Q \simeq \hom(Q, \mathbb{Z})}$.

For example, the free part of the middle cohomology group of an oriented manifold of dimension divisible by four is a lattice. Moreover, ${K}$ is a lattice.

Given an element ${x \in K}$, we know that we can represent ${x}$ by a map ${\alpha: S^m \rightarrow M}$ whose composite ${S^m \rightarrow M \rightarrow X}$ is nullhomotopic. Using a strong form of the Whitney imbedding theorem, we can assume that ${\alpha}$ is an imbedding—here we need ${m \geq 3}$. In particular, the quadratic form evaluated on ${x}$, that is ${(x,x)}$, is the self-intersection of ${S^m}$ in ${M}$.

Lemma 10 Suppose ${m}$ is even. Then ${(x,x) \in 2 \mathbb{Z}}$ for ${x \in K}$. That is, any imbedding ${\alpha: S^m \hookrightarrow M}$ whose composite ${S^m \rightarrow M \rightarrow X}$ is trivial has even self-intersection. Moreover, ${(x,x) =0 }$ if and only if the normal bundle of ${\alpha}$ is trivial.

Proof: In fact, the self-intersection of ${S^m}$ with itself inside ${M}$ is the Euler class of the normal bundle to ${S^m}$ inside ${M}$. We know that the normal bundle, which is classified by an element ${t \in \pi_{m-1}(SO(m))}$, isstably trivial because ${S^m \stackrel{\alpha}{\rightarrow} M \rightarrow X}$ is nullhomotopic and because ${ M \rightarrow X}$ is a normal map. It follows that ${t}$ is killed under the map

$\displaystyle \pi_{m-1}(SO(m)) \rightarrow \pi_{m-1}(SO(m+1)) \simeq \pi_{m-1}(SO).$

Via the fibration ${SO(m) \rightarrow SO(m+1) \rightarrow S^m}$, we find that ${t}$ is in the image of the boundary map

$\displaystyle \pi_m(S^m) \rightarrow \pi_{m-1}(SO(m)),$

which sends ${1 \in \pi_m(S^m)}$ to the class of the tangent bundle of ${S^m}$. Consequently, the Euler class ${(x,x)}$ of ${t}$ (the normal bundle of ${S^m \stackrel{\alpha}{\hookrightarrow} M}$) is a multiple of the Euler characteristic of ${S^m}$, so ${(x,x) \in 2 \mathbb{Z}}$. This multiple is trivial if and only if ${(x,x) = 0}$. $\Box$

The previous lemma establishes the connection between the algebra of the bilinear pairing on ${K}$ and the geometry of the maps ${\alpha: S^m \rightarrow M}$. We now use the following fact about quadratic forms:

Lemma 11 Let ${K}$ be a lattice of dimension ${d}$. Suppose ${b(x,x) \in 2 \mathbb{Z}}$ for ${x \in K}$, and further that the signature of ${b}$ (on ${K \otimes_{\mathbb{Z}} \mathbb{R}}$) is zero. Then ${(K, b)}$ is isomorphic to a direct sum of hyperbolic lattices; in particular ${d}$ is even. That is, there exists a basis ${\{x_1, \dots, x_{d/2}, y_1, \dots, y_{d/2}\}}$ of ${K}$ such that$\displaystyle (x_i, x_j) = (y_i, y_j) = 0, \quad (x_i, y_j) = \delta_{ij}.$

See this paper by Milnor, lemma 9.

This lemma on quadratic forms allows us to do surgery to kill ${K}$. Namely, choose a basis as above of ${K}$, and let us explain how to kill ${\mathbb{Z}x_1 \oplus \mathbb{Z}y_1}$, for instance. Using the strong form of the Whitney imbedding theorem (see the same paper by Milnor), choose an imbedding

$\displaystyle \alpha: S^m \rightarrow M$

representing ${x_1}$ (or rather, its Poincaré dual), as well as a nullhomotopy of ${f \circ \alpha: S^m \rightarrow M \rightarrow X}$. Since ${(x_1, x_1) = 0}$, we find that the normal bundle of ${\alpha}$ is trivial.

In particular, we can choose an imbedding ${S^m \times D^m \hookrightarrow M}$ extending ${\alpha}$. Such an imbedding is equivalent (by the tubular neighborhood theorem) to a choice of trivialization of the normal bundle. The trivializations of the normal bundle of ${\alpha}$ are parametrized by ${\pi_m(SO(m))}$, while the trivializations of the stable normal bundle are parametrized by ${\pi_m(SO)}$. Now, the map

$\displaystyle \pi_m(SO(m)) \rightarrow \pi_m(SO(m+1)) \twoheadrightarrow \pi_m(SO)$

is a surjection. In fact, the cokernel of the first map injects into ${\mathrm{ker}( \pi_m(S^m) \rightarrow \pi_{m-1}(SO(m))}$, which is trivial since it sends the generator to the class tangent bundle of ${S^m}$ (whose Euler class is nontrivial). Therefore, we can choose a trivialization of the normal bundle of ${\alpha}$ which is compatible with the trivialization of ${f^* \xi}$ given by the nullhomotopy of ${f \circ \alpha: S^n \rightarrow X}$.

In particular, we can do surgery, just as we could before, to get a map

$\displaystyle f': M' \rightarrow X, \quad M' = M \setminus (S^m \times \mathrm{Int} D^m) \cup_{S^m \times S^{m-1}} (D^{m+1} \times S^{m-1})$

It is not obvious that the map ${f'}$ is still ${m}$-connected and that we have actually simplified the relative homotopy groups. To see this, we consider the intermediate manifold-with-boundary ${M_0 = M \setminus (S^m \times \mathrm{Int}(D^m))}$. This is homotopy equivalent to ${M \setminus (S^m \times \left\{0\right\})}$ and to ${M' \setminus ( \left\{0\right\} \times S^{m-1})}$, which interpolates between ${M}$ and ${M'}$.

Fix a field ${k}$ and take coefficients in ${k}$. We have a long exact sequence in cohomology

$\displaystyle \dots \rightarrow H^{m-1}(M) \rightarrow H^{m-1}(M_0) \rightarrow H^m(M, M_0) \rightarrow H^m(M) \rightarrow H^m(M_0) \rightarrow \dots.$

Observe that ${H^m(M, M_0) \simeq H^m(S^m \times D^m, S^m \times (D^m \setminus \left\{0\right\})) \simeq k}$ and the lower-dimensional cohomology groups of ${M, M_0}$ vanish. The map

$\displaystyle H^m(M, M_0) \simeq k \rightarrow H^m(M)$

can be identified with the map ${1 \mapsto x_1}$. We find that it is an injection. It follows that:

1. ${H^i(M) \rightarrow H^i(M_0)}$ is an isomorphism for ${i < m}$.
2. There is an exact sequence$\displaystyle H^m(M, M_0) \rightarrow H^m(M) \rightarrow H^m(M_0) \rightarrow 0,$

which shows that ${H^m(M_0)}$ is a quotient space of ${H_m(M)}$, by the image of ${x_1}$.

We now play the same game with ${M_0 \rightarrow M'}$. Observe that the inclusion ${M_0 \rightarrow M'}$ is homotopy equivalent to ${M' \setminus ( \left\{0\right\} \times S^{m-1}) \rightarrow M'}$, as observed above. Consequently,

$\displaystyle H^*(M', M_0) \simeq H^*( D^{m+1} \times S^{m-1}, (D^{m+1} \setminus \left\{0\right\}) \times S^{m-1})).$

This gives an exact sequence

$\displaystyle \dots \rightarrow 0 \rightarrow H^m(M') \rightarrow H^m(M_0) \rightarrow H^{m+1}(M', M_0) \simeq k ,$

which shows that ${H^i(M_0) \rightarrow H^i(M')}$ is an isomorphism for ${i < m}$. Moreover, we find that ${H^m(M') \subset H^m(M_0) }$, so it is a ${k}$-vector space is of dimension at most that of ${ H^m(M_0)/k x_1}$. In fact, the commutative diagram:

shows that the map ${H^m(M_0) \rightarrow H^{m+1}(M', M_0)}$ can be identified with pull-back along ${\alpha: S^m \rightarrow M_0}$. In particular, it maps ${y_1}$ to ${1}$ and consequently ${H^m(M)}$ is strictly smaller than ${H^m(M_0)}$. (We don’t actually need to know this, though.)

What does this tell us? Starting with a degree one normal map ${f: M \rightarrow X}$ which was ${m}$-connected, we have produced (by surgery on ${f}$) a new degree one normal map ${f': M' \rightarrow X}$ with the following properties.

1. The rank of ${H_m(M')}$ is smaller than that of ${H_m(M)}$. In particular, the kernel ${K(f')}$ is smaller than ${K(f)}$.
2. ${f': M' \rightarrow X}$ is still ${m}$-connected. To see this, observe that we had a zig-zag of isomorphisms between ${H_i(M)}$ and ${H_i(M')}$ for ${i < m}$. Namely, we have a commutative diagram:

where the map ${f_*: H_i(M) \rightarrow H_i(X)}$ is an isomorphism because ${f}$ is ${m}$-connected. This diagram shows that ${f'_*: H_i(M') \rightarrow H_i(X)}$ is an isomorphism for ${i < m}$. For ${i = m}$, the map ${f'_*}$ is already a surjection because ${f'}$ is a normal map.

Repeating this process, we can keep applying surgery to get a manifold homotopy equivalent to ${X}$.