Proof. Because the cohomology groups $H^p(B,\mathbb {Z})$ are free, the $E_2$ page of the spectral sequence can be identified with

(2.5) $$ \begin{align} E_2^{p,q}= H^p(B)\otimes \mathbb{E}^q(pt). \end{align} $$

The spectral sequence is linear over $\mathbb {E}^q(pt)$ and it suffices to prove that

(2.6) $$ \begin{align} d^r(\alpha\otimes 1)=0 \text{ for all } \alpha \in H^*(B). \end{align} $$

If $\mathbb {E}^*=MU^*$ is complex cobordism the spectral sequence collapses because the $E_2$ page is concentrated in even degrees. In general, the complex orientation of $\mathbb {E}^*$ is induced by a natural transformation (of multiplicative cohomology theories) $MU^* => \mathbb {E}^*$ and the Atiyah-Hirzebruch sequence is natural in the cohomology theory. The result therefore follows from the case of complex cobordism.

Proof. Suppose that we have a Hamiltonian fibration $M \xrightarrow {i} P \xrightarrow {\pi } B$ over B and let $M \xrightarrow {\tilde {i}} \tilde {P} \xrightarrow {\tilde {\pi }} \tilde {B}$ be the pull-back Hamiltonian fibration, which admits a map $\tau ^P:\tilde {P} \to P.$ We can factor $\tau ^B$ as an inclusion followed by a projection $\tilde {B} \to \mathbb {CP}^r \times B \to B$ and similarly for $\tau ^P$ . This shows that the map $\tau ^P$ is compatible with complex-orientations. Set $U \subset B$ to be the locus over which $\tau ^{B}$ is an isomorphism and let $P_U$ denote the restriction of the bundle to this locus. Note that we can assume that our reference fiber $M_b$ for both $\tilde {P}$ and P lies over a point $b \in U$ . We have a Cartesian diagram of complex-oriented manifolds/maps:

For such Cartesian squares, the Umkëhr maps on cohomology are compatible with pullback/restriction maps; see [Reference QuillenQui71, 1.7] for complex cobordism, [Reference Bressler and EvensBE90, E(iii)] in general, and [Reference Crabb and JamesCJ98, Proposition 12.9] for a statement at the level of spectra, which implies the statement for cohomology by taking homotopy groups. In the present case this means that:

(2.7) $$ \begin{align} j^* \circ \tau^P_*=\tilde{j^*} \text{ and hence } i^* \circ \tau^P_* =\tilde{i}^*. \end{align} $$

Suppose we are given a splitting $\tilde {s}:\mathbb {E}^*(M) \to \mathbb {E}^*(\tilde {P})$ such that $(\tilde {i})^* \circ \tilde {s}= \mathrm {id}$ . Then it follows from (2.7) that

(2.8) $$ \begin{align} s=\tau^P_* \circ \tilde{s}:\mathbb{E}^*(M) \to \mathbb{E}^*(P) \end{align} $$

satisfies $i^* \circ s=\mathrm {id}.$

Proof. Let $F \to E \to P$ be a Hamiltonian bundle over P. After restricting to M we get a Hamiltonian bundle $F \to W \to M$ which by assumption gives rise to a c-splitting:

(2.9) $$ \begin{align} s_M: \mathbb{E}^*(F) \to \mathbb{E}^*(W). \end{align} $$

By [Reference Lalonde and McDuffLM03, Lemma 3.9], the composite bundle $ W \to E \to B$ is also Hamiltonian, thus gives rise to a c-splitting:

(2.10) $$ \begin{align} s_B: \mathbb{E}^*(W) \to\mathbb{E}^*(E). \end{align} $$

The desired c-splitting for $F \to E \to P$ is the map $s=s_B \circ s_M$ .

Proof. In view of Lemma 2.7, it suffices to consider the case where $B=\mathbb {CP}^{k}$ is a single projective space. We proceed by induction on k. The base case where $k=1$ is Theorem 2.4. Consider the blow-up $\tilde {B}$ of $\mathbb {CP}^{k}$ at a point. Then $\tilde {B}$ can be identified with the $\mathbb {CP}^1$ -bundle:

(2.11) $$ \begin{align} \mathbb{P}(\mathcal{O}_{\mathbb{CP}^{k-1}}\oplus \mathcal{O}_{\mathbb{CP}^{k-1}}(-1)) \to \mathbb{CP}^{k-1}. \end{align} $$

The inductive hypothesis together with Lemma 2.7 shows that $\tilde {B}$ is Hamiltonian c-split. It then follows from Lemma 2.6 that $\mathbb {CP}^{k}$ is c-split.

Proof. Again, it suffices to consider a single Grassmannian. Recall that the Schubert varieties of the complete flag variety $GL(n,{\mathbb C})/B$ are indexed by elements w of the symmetric group $S_n$ . If we consider the Schubert variety $Z_w$ corresponding to the permutation $w(i) = i+k (\mathrm {mod} n)$ , this maps birationally onto the complex Grassmannian $\mathrm {Gr}(k,n)$ via the projection $GL(n,{\mathbb C})/B \to \mathrm {Gr}(k,n)$ (see discussion above Example 1.2.3 in [Reference BrionBri05]). Now take the Bott–Samelson variety $Z_{\underline {w}}$ associated to a reduced decomposition $\underline {w}$ of w. By, for example, [Reference BrionBri05, Proposition 2.2.1 (iii), (iv)], this is an iterated fibration of Kähler manifolds

(2.12) $$ \begin{align} M_k \to M_{k-1} \to \cdots \to M_1=\mathbb{CP}^1 \end{align} $$

with fiber $\mathbb {CP}^1$ , which maps birationally onto $Z_w$ and hence onto $\mathrm {Gr}(k,n)$ . The statement again follows from Lemma 2.6 and Lemma 2.7.

Proof. Part (1): We follow the proof of the Leray–Hirsch theorem in [Reference HatcherHat02, Theorem 4D.1]. We assume that B has dimension m and proceed by induction on m. We let $B'$ be the subspace obtained by deleting a point $x_\alpha $ from each m cell $U_\alpha $ and let $P'$ denote the restriction of the fibration to $B'$ . Then $M \to P' \to B'$ is also a c-split fibration. We let $V_\alpha \subset U_\alpha $ be a small disc about $x_\alpha $ over which the fibration is trivial.

We note that the cohomologies $\mathbb {E}^*(B,B')$ , $\mathbb {E}^*(B)$ , $\mathbb {E}^*(B')$ are all free over $\mathbb {E}_*$ . For $\mathbb {E}^*(B)$ , $\mathbb {E}^*(B')$ this follows from Lemma 2.1, together with the fact that $B'$ retracts onto the $m-2$ skeleton. For $\mathbb {E}^*(B,B')$ , this follows from the excision isomorphism $\mathbb {E}^*(B,B') \cong \mathbb {E}^*(V_\alpha ,V_\alpha -x_\alpha ).$ Lemma 2.1 implies that the maps $\mathbb {E}^*(B) \to \mathbb {E}^*(B')$ are surjections as the assertion holds for the integral cohomology. It follows from the long exact sequence for the pair $(B,B')$ that the connecting maps

(2.15) $$ \begin{align} \delta :\mathbb{E}^*(B') \to \mathbb{E}^{*+1}(B,B') \text{ are zero.} \end{align} $$

This means that the long exact sequence gives rise to a short exact sequence:

(2.16) $$ \begin{align} 0 \to \mathbb{E}^*(B,B') \to \mathbb{E}^*(B) \to \mathbb{E}^*(B') \to 0 \end{align} $$

Because all of these groups are finite free over $\mathbb {E}_*$ , we learn that

(2.17) $$ \begin{align} 0 \to \mathbb{E}^*(B,B')\otimes_{\mathbb{E}_{*}}\mathbb{E}^*(M) \to \mathbb{E}^*(B)\otimes_{\mathbb{E}_{*}}\mathbb{E}^*(M) \to \mathbb{E}^*(B')\otimes_{\mathbb{E}_{*}}\mathbb{E}^*(M) \to 0 \end{align} $$

is also exact. Let $(\mathbb {E}^*(B,B')\otimes _{\mathbb {E}_{*}}\mathbb {E}^*(M))^{q}$ denote the degree q piece of the tensor product and similarly for the other groups. Then there is a commutative diagram:

The top row is exact by (2.17) and (2.15) and the commutativity of the diagram follows as in the proof of [Reference HatcherHat02, Theorem 4D.1]. It is then easy to establish that $\phi : \mathbb {E}^*(B,B')\otimes _{\mathbb {E}_{*}}\mathbb {E}^*(M) \to \mathbb {E}^*(P,P')$ is an isomorphism by retracting everything on to $V_\alpha .$ Because $B'$ retracts onto the $m-2$ skeleton, the induction hypothesis implies that the map $\phi : \mathbb {E}^*(B')\otimes _{\mathbb {E}_{*}}\mathbb {E}^*(M) \to \mathbb {E}^*(P')$ is also an isomorphism. The claim now follows from the five-lemma.

Part (2): We let $B^m$ denote the m-skeleton of B and let $P^m \to B^m$ denote the restriction of P to $B^m$ . Using the splitting s, we can identify

(2.18) $$ \begin{align} \phi^m: \mathbb{E}^*(M)\otimes_{\mathbb{E}_*} \mathbb{E}^*(B^m) \to \mathbb{E}^*(P^m) \end{align} $$

In particular, all of the restriction maps $\mathbb {E}^*(P^{m+1}) \to \mathbb {E}^*(P^m)$ are surjective. Applying the Milnor exact sequence together with the Mittag-Leffler property, we obtain isomorphisms:

(2.19) $$ \begin{align} \varprojlim_m \mathbb{E}^*(M)\otimes_{\mathbb{E}_*} \mathbb{E}^*(B^m) \cong \varprojlim_m \mathbb{E}^*(P^m) \cong \mathbb{E}^*(P). \end{align} $$

Finally, we note that the same argument shows that the inverse system $\varprojlim _m \mathbb {E}^*(M)\otimes _{\mathbb {E}_*} \mathbb {E}^*(B^m)$ is canonically isomorphic to $\mathbb {E}^*(M \times B)$ , proving the claim.

Proof. This follows immediately by combining Lemma 2.9 and Lemma 2.11.

Proof. Let $\tilde {s}$ be the c-splitting over B. This gives rise to a commutative diagram:

where the horizontal arrows are isomorphisms. Thus, because $\mathbb {E}^*(B)$ , $\mathbb {E}^*(B_0)$ are free, we may choose a splitting of $\mathbb {E}_*$ modules $\mathbb {E}^*(B_0)\to \mathbb {E}^*(B)$ for $i^*$ . This defines a splitting $\tau : \mathbb {E}^*(P_{0})\hookrightarrow \mathbb {E}^*(P)$ of $\mathbb {E}_*$ -modules for $\iota ^*.$ We can thus split the cohomology $\mathbb {E}^*(P)$ as

(2.22) $$ \begin{align}\mathbb{E}^*(P) \cong \tau(\mathbb{E}^*(P_{0}))\oplus \mathrm{ker}(\iota^*)\end{align} $$

We then define a new splitting $s:\mathbb {E}^*(M) \to \mathbb {E}^*(P)$ using $s_{0}$ by letting $s=(\tau \circ s_0,0)$ in this direct sum decomposition. By construction s satisfies (2.21).

Proof of Theorem 1.2

Part (1): Recall that there is a convenient model for the total space of the universal $U(k)$ -bundle, $EU(k)$ , given by taking the set of orthonormal k-frames in a complex Hilbert space $\mathcal {H}.$ Topologically, this is the colimit of spaces of k-frames in $\mathbb {C}^n$ as $n \to \infty $ :

$$ \begin{align*}EU(k)_1 \subset \cdots \subset EU(k)_n \subset \cdots .\end{align*} $$

The group $U(k)$ acts freely on this space and the quotient is the infinite Grassmannian of k-planes $Gr_k(\mathcal {H}).$ Given a product of unitary groups $G=\prod U(k_i)$ , we can similarly take $EG:=\prod EU(k_i).$ The classifying space for G, $BG$ , therefore admits finite dimensional approximations by spaces $BG_n$ which are products of Grassmannians. We set $M_G:=M\times _G EG$ and $M_{G,n}:=M\times _G EG_n$ . In view of Lemma 2.9, we have c-splittings $s_n: \mathbb {E}^*(M) \to \mathbb {E}^*(M_{G,n})$ which by Corollary 2.13 give rise to identifications:

(2.23) $$ \begin{align} \mathbb{E}^*(M_{G,n}) \cong \mathbb{E}^*(M)\otimes \mathbb{E}^*( BG_n).\end{align} $$

The identifications from (2.23) imply that the restrictions $\mathbb {E}^*(M_{G,n+1}) \to \mathbb {E}^*(M_{G,n})$ are surjective and we have that

(2.24) $$ \begin{align}\mathbb{E}^*(M_G) \cong \varprojlim_n\mathbb{E}^*(M_{G,n}). \end{align} $$

By Lemma 2.14, we can arrange that the c-splittings $s_n$ are compatible with restriction and thus give rise to a splitting:

(2.25) $$ \begin{align}s: \mathbb{E}^*(M) \to \mathbb{E}^*(M_G). \end{align} $$

Using Lemma 2.11 (2) (by construction the model for $BG$ has only even-dimensional cells), we obtain that

(2.26) $$ \begin{align} \mathbb{E}^*(M_G) \cong \mathbb{E}^*(M \times BG). \end{align} $$

It remains to prove that the Leray-spectral sequence $\lbrace E_r^{p,q}(M_{G}) \rbrace $ for the fibration $M_G \to BG$ degenerates. It follows easily from (2.23) that the Leray spectral sequences $\lbrace E_r^{p,q}(M_{G,n}) \rbrace $ associated to the fibrations $M_{G,n} \to BG_n$ degenerate. The $E_2$ page $E_2^{p,q}(M_{G})$ is an inverse limit of the $E_2$ pages $E_2^{p,q}(M_{G,n})$ , which implies that the differential vanishes on $E_2^{p,q}(M_{G})$ . The same argument then implies that the differentials on the subsequent pages $E_r^{p,q}(M_{G})$ are zero, which proves degeneration.

Part (2): We let $EG$ be any CW-complex on which G (and hence $T \subset G$ ) acts freely. We again set $M_T=M\times _T EG$ and $M_G:=M\times _G EG$ and consider the diagram

We let $\lbrace E_r^{p,q}(M_T) \rbrace $ and $\lbrace E_r^{p,q}(M_G) \rbrace $ be the Leray spectral sequences associated to the fibrations over the respective Borel spaces $BT$ and $BG$ . By functoriality of the Leray spectral sequence, r induces maps $r^*: E_r(M_G) \to E_r(M_T).$ We can identify the $E_2$ pages of the spectral sequence with $E_2(M_G)=\mathbb {E}^*(M)\otimes _{\mathbf {k}}H^*(BG,\mathbf {k})$ and $E_2(M_T)=\mathbb {E}^*(M)\otimes _{\mathbf {k}}H^*(BT,\mathbf {k})$ respectively. By hypothesis $r^*$ is injective on the $E_2$ pages and hence there are no differentials on $E_2(M_G).$ The same argument shows there are therefore no differentials on subsequent pages and thus the spectral sequence $E_r^{p,q}(M_G)=>\mathbb {E}^*(M_G)$ degenerates at the second page as required.

Proof. By equivariant formality, $\mathbb {E}^*_T(M) \cong \mathbb {E}^*(M\times BT).$ We also have an identification $\mathbb {E}^*(M\times BT) \cong \mathbb {E}^*(M)[\![u_1,\cdots ,u_m]\!]$ from which the result follows.

Proof. Part (1) is established in [Reference Hopkins, Kuhn and RavenelHKR00, Cor. 5.8], so we focus on explaining Part (2). Note that by Part (1), we have that the pull-back map

(3.5) $$ \begin{align} \mathbb{E}^*(BA) \to \mathbb{E}^*(B\mathbb{Z}/p^r\mathbb{Z}) \end{align} $$

is an isomorphism. Thus, it suffices to consider the case where $\ell =p^r.$ View A as a finite subgroup $A \subset S^1$ . Regard $\mathbb {E}^*(BA)$ as a module over $\mathbb {E}_*[\![u]\!]$ via the map $BA \to BS^1$ . By [Reference Hopkins, Kuhn and RavenelHKR00, Lemma 5.7] we have that

(3.6) $$ \begin{align} \mathbb{E}^*(BA) \cong \mathbb{E}_*[\![u]\!]/([p^r]\cdot_{\mathbb{E}}u). \end{align} $$

The formal group law $L_{\mathbb {E}}$ for Morava K-theory is essentially characterized by the fact that its p-series satisfies:

(3.7) $$ \begin{align} [p]\cdot_{\mathbb{E}}u = v_n u^{p^{n}}.\end{align} $$

Iterating the formal group law, we obtain

(3.8) $$ \begin{align} [p^r] \cdot_{\mathbb{E}} u =v_n^{d(r)} u^{p^{rn}}. \end{align} $$

for some integer $d(r)>0$ . The result follows immediately from this.

Proof. We have that $\mathbb {E}_{T}^*(X) \cong \mathbb {E}^*(BK)$ and the induced $\mathbb {E}^*(BT)$ module structure on this group is induced by the inclusion $K \subset T$ . We first argue that the map

(3.9) $$ \begin{align}\mathbb{E}^*(BT) \to \mathbb{E}^*(BK) \end{align} $$

is surjective so that $\mathrm {Spec}(\mathbb {E}^*(BK))$ is a closed subscheme of $\mathrm {Spec}(\mathbb {E}^*(BT))$ . As above, we let $K^0$ denote the connected component of K. After choosing a complementary torus H to $K^0$ , we may write $K=A \times K^0$ , where A is a finite abelian subgroup of H. Then by the Künneth isomorphism [Reference Hopkins, Kuhn and RavenelHKR00, Cor. 5.11], we have

(3.10) $$ \begin{align} \mathbb{E}^*(BA)\otimes \mathbb{E}^*(BK^0) \cong \mathbb{E}^*(BK). \end{align} $$

Moreover, the Kunneth map fits into a diagram:

It therefore suffices to assume that $K=A$ is a finite abelian group. Moreover, by Part (1) of Lemma 3.2 we can assume that $A \cong \mathbb {Z}/p^{r_1} \times \mathbb {Z}/p^{r_2} \times \cdots \times \mathbb {Z}/p^{r_i}$ is a product of p-groups. It is easy to see that we may find:

• ○ a rank one subgroup $H' \cong S^1 \subset T$ so that $\mathbb {Z}/p^{r_1} \subset H'$

• ○ a complementary torus $H"$ (for $H'$ ) together with a subgroup $A' \subset H"$ so that $A \cong \mathbb {Z}/p^{r_1} \times A'$ (the group $A' \cong \mathbb {Z}/p^{r_2} \times \cdots \times \mathbb {Z}/p^{r_i}$ ).

The surjectivity of the restriction $\mathbb {E}^*(BT) \to \mathbb {E}^*(BA)$ follows from the Kunneth isomorphism together with induction on the number of factors in the product decomposition of A. To prove the claim on dimensions, we note that the claim is obvious for connected groups where $K=K^0$ . In view of the Kunneth isomorphism and Part (1) of Lemma 3.2, we have that $\mathbb {E}^*(BK)$ is finite over $\mathbb {E}^*(BK^0)$ , which implies that they have equal dimensions [Sta24, Tag 01WG]. To prove Part (2) of the Lemma, we again write our group K as $A\times K^0$ . Without loss of generality, we can assume $A \cong \mathbb {Z}/p^{r_1} \times \mathbb {Z}/p^{r_2} \times \cdots \times \mathbb {Z}/p^{r_i}$ . Then Part (2) of Lemma 3.2 shows that $K(n)^*(BA)$ is a tensor product of the algebras in (3.4). In particular, $K(n)^*(BA)$ is a nilpotent extension of $K(n)_*$ for any finite abelian group. The claim now follows from the Kunneth isomorphism (3.10).

Proof. We stratify the complement of an equivariant tubular neighborhood U of $M^T$ by finitely many open sets which equivariantly retract onto nonconstant orbits. Note that for a nonconstant orbit X, Lemma 3.5 implies in particular that

(3.12) $$ \begin{align} \mathbb{E}_{T}^*(X)\otimes_{\mathbb{E}^*(BT)} R =0. \end{align} $$

The result then follows from Lemma 3.2, Lemma 3.5, and the Mayer-Vietoris sequence.

Proof of Theorem 1.10

The assumption that $\mathbb {E}_*$ is free implies by Corollary 2.16 that

(3.13) $$ \begin{align} \mathbb{E}_T^*(M) \cong \mathbb{E}^*(M)[[u_1,\cdots,u_m]] \cong \mathbb{E}^*(M)\otimes_{\mathbb{E}_*} \mathbb{E}^*(BT). \end{align} $$

Hence, $\mathbb {E}_T^*(M)$ is free as a module over $\mathbb {E}^*(BT).$ By Lemma 3.6, the kernel of $j^*$ is a torsion module over $\mathbb {E}^*_T(pt)$ . Because $\mathbb {E}^*_{T}(M)$ is free, it follows that the kernel of $j^*$ is trivial.

Proof. By the complex-orientation assumption and its compatibility with the T-action, we see that the pushforward in equivariant cohomology $j_*$ and the Euler class of the normal bundle $\nu _i$ are well-defined. For each $i \in I$ , we have $j_i^* j_* (1) = \mathit {eu}^{T}_{\mathbb {E}}(\nu _i)$ . Using the Mayer–Vietoris long exact sequence for the pair $(M, M \setminus M^T)$ , as in Lemma 3.6, one can show that for the pushforward $j_*$ , both $\ker (j_*)$ and $\mathrm {coker}(j_*)$ are torsion modules over $\mathbb {E}^*(BT)$ . Combining with Lemma 3.6, it follows that $j^* j_*$ is an isomorphism. This proves the invertibility of the Euler classes, so the statement follows accordingly.

Proof. The fact that $\ker (j)^*$ is trivial follows immediately from Theorem 1.10 because the coefficient ring is a graded field. Because the connecting homomorphism $\delta $ factors as the composition

(3.20) $$ \begin{align} \mathbb{E}^*_{T}(M^T) \xrightarrow{\delta'} \mathbb{E}^{*+1}_T(M, M^T) \rightarrow \mathbb{E}^{*+1}_T(M_1, M^T) \end{align} $$

where recall that $\delta '$ is the connecting homomorphism for the pair $(M,M^T)$ , the long exact sequence implies that $\delta \circ j^* = 0$ . To finish the proof, suppose $\xi \in \mathbb {E}^*_{T}(M^T)$ satisfies $\delta (\xi ) = 0$ . To show it lies in the image of $j^*$ , we need to show that $\delta '(\xi ) = 0$ .

Suppose this is not the case. We claim that the ideal $I(\xi )$ is a principal ideal. We follow the proof of [Reference Chang and SkjelbredCS74, Lemma 2.2]. Indeed, let $0 \neq a \in I(\xi )$ be an element which is not divisible by any other element and suppose $b \in I(\xi )$ . Choosing a basis $\{x_i\} \subset \mathbb {E}^*_T(M)$ , we can find $a_i, b_i \in \mathbb {E}^*_{T}(pt)$ such that $a\xi = j^*(\sum a_i x_i)$ and $b \xi = j^*(\sum b_i x_i)$ using the long exact sequence for the pair $(M, M^T)$ . Then we know that $j^*(\sum b a_i x_i) = j^*(\sum a b_i x_i)$ , so $\sum b a_i x_i = \sum a b_i x_i$ by the injectivity of $j^*$ . This shows that for each i, we have $b a_i = a b_i$ . As the $a_i$ must have no common factor, we see that b is divisible by a, which proves the claim.

By Lemma 3.9, as $I(\xi )$ is principal, there must be one $\mathfrak {m}$ which has codimension $1$ in T. Because of the inclusion $M^{\mathfrak {m}} \subset M_1$ , the connecting homomorphism $\delta ^K$ factors as

(3.21) $$ \begin{align} \mathbb{E}^*_{T}(M^T) \xrightarrow{\delta} \mathbb{E}^{*+1}_T(M_1, M^T) \to \mathbb{E}^{*+1}_T(M^{\mathfrak{m}}, M^T), \end{align} $$

which tells us that $\delta (\xi ) \neq 0$ , leading to a contradiction.

Proof of Theorem 1.12

This follows from the proof of [Reference Goresky, Kottwitz and MacPhersonGKM98, Theorem 7.2], which we reproduce for completeness. Because $\mathbb {E}^*_T(M)$ is a free module over $\mathbb {E}^*(BT)$ , using the exact sequence (3.19), it suffices to determine the kernel of $\delta : \mathbb {E}^*_T(M^T) \to \mathbb {E}^{*+1}_T(M_1, M^T)$ . Given the complement of poles $E_j$ , by definition, its stabilizer group is $\ker (\Theta _j)$ . Therefore,

(3.22) $$ \begin{align} \mathbb{E}^*_{T}(E_j) \cong \mathbb{E}^*_{T}(T / \ker(\Theta_j)) \cong \mathbb{E}^*_{\ker(\Theta_j)}(pt). \end{align} $$

Choose a T-invariant open cover of $\overline {E}_j = U_j \cup V_j$ where $U_j = \overline {E}_j \setminus \{ x_{j_0} \}$ and $V_j = \overline {E}_j \setminus \{ x_{j_{\infty }} \}$ . Then the corresponding Mayer–Vietoris long exact sequence splits into short exact sequences because all the cohomology groups concentrate in even degrees. Furthermore, we have a commutative diagram

(3.23)

which tells us that $\delta $ agrees with the map

(3.24) $$ \begin{align} (f_{j_0}, f_{j_{\infty}}) \mapsto f_{j_0} |_{\mathbb{E}^*_{\ker(\Theta_j)}} - f_{j_{\infty}} |_{\mathbb{E}^*_{\ker(\Theta_j)}}. \end{align} $$

Ranging over all j, the theorem is proved.