2. Construction of the map $t:\mathcal {N}(X^{n}) \to H^{st}_{n} ( X^{n}; \mathbb {L}^{+})$

We recall the construction of the map $t:\mathcal {N}(X^{n}) \to H^{st}_{n} ( X^{n}; \mathbb {L}^{+})$ from Hegenbarth and Repovš [Reference Hegenbarth and Repovš9, Section 4]. So let us fix an oriented closed generalized $n$-manifold $X^{n}$ of dimension $n \ge 5$. If $\mathcal {U}$ is a covering of $X^{n}$ by open sets, we denote its nerve by $N(\mathcal {U}).$ If the covering $\mathcal {U'} \prec \mathcal {U}$ is a refinement of $\mathcal {U}$, then there is a simplicial map $s: N(\mathcal {U'}) \to N(\mathcal {U}).$

Let $M(s_{j})= N (\mathcal {U}_{j+1}) \times I \underset {s_j}{\cup } N (\mathcal {U}_{j})$ be the mapping cylinder of the map $s_{j}:N(\mathcal {U}_{j+1}) \to N(\mathcal {U}_{j}).$ Using property Proposition 2.1 (d), we can form the mapping telescope $F_{0}= \underset {j \in \mathbb {N}}{\cup } M(s_{j})$ and the obvious maps

\[ X^{n} \times [j, j+1] \xrightarrow[]{\varphi_{j} \times Id_{[j, j+1]}} M(s_{j}) \xrightarrow[]{\psi_{j} \times Id_{[j, j+1]}} X^{n} \times [j, j+1] \]

fit together to give the map $X^{n}\times \mathbb {R}_{+} \xrightarrow []{\varGamma } F_{0} \xrightarrow []{\Lambda } X^{n}\times \mathbb {R}_{+}.$

Here, $F_{0}$ is a locally finite complex which can be completed to give a complex $F$ such that (cf. Hegenbarth and Repovš [Reference Hegenbarth and Repovš9, Section 3] for details):

1. (i) at the $\infty$-end, we add

   \[ \underset{j}{\varprojlim} N({\mathcal{U}}_{j}) = X^{n}; \]

2. (ii) at the $0$-end, we add a cone with the cone point $c_{0}.$

The complex $F_{0}$ (respectively $F$) is an open (respectively closed) fundamental complex of the (compact metric) space $X^{n}$. If $\mathbb {E}$ is an arbitrary spectrum and $H^{lf}_{*}(F_{0}; \mathbb {E})$ denotes the locally finite homology of $F_{0}$, then the Steenrod homology satisfies the following axiom

\[ H^{lf}_{*}(F_{0}; \mathbb{E}) \cong H^{st}_{*}(F, X^{n},\{c_{0}\}; \mathbb{E}). \]

Note that $F$ is contractible, hence we have the following isomorphism

\[ H^{st}_{m}(F, X^{n},\{c_{0}\}; \mathbb{E}) \xrightarrow[{\cong}]{\partial} H^{st}_{m-1}(X^{n}; \mathbb{E}). \]

We can now outline the construction of the map $t:\mathcal {N}(X^{n}) \to H^{st}_{n} ( X^{n}; \mathbb {L}^{+})$ (cf. Hegenbarth and Repovš [Reference Hegenbarth and Repovš9, Section 4]). Let $(f,\,b)$ be a normal degree one map, i.e. $f:M^{n} \to X^{n}$ is of degree one and $b:\nu _{M^{n}} \to \xi$ is a bundle map covering $f$. As before, $(f_{0},\,b_{0})$ denotes the canonical map, i.e. $f_{0}:M^{n}_{0} \to X^{n},\, \ b_{0}:\nu _{M^{n}_{0}} \to \xi _{0}.$ Consider the following bundles over $F_{0}$: $\eta = \Lambda ^{*}(\xi \times \mathbb {R}_{+}),\, \ \eta _{0} = \Lambda ^{*}(\xi _{0} \times \mathbb {R}_{+}).$ Then $\varGamma ^{*}(\eta )\cong \xi \times \mathbb {R}_{+},\, \ \varGamma ^{*}(\eta _{0})\cong \xi _{0} \times \mathbb {R}_{+},$ since $\Lambda \circ \varGamma$ is homotopic to $Id_{X^{n}\times \mathbb {R}_{+} }$.

One obtains bundle maps $(\Phi ,\, B)$ and $(\Phi _{0},\, B_{0})$ from the following compositions

\begin{align*} & \Phi: M^{n} \times \mathbb{R}_{+} \xrightarrow[]{f \times Id_{\mathbb{R}_{+}}} X^{n} \times \mathbb{R}_{+} \xrightarrow[]{\varGamma} F_{0},\\ & B: \nu_{M^{n}} \times \mathbb{R}_{+} \xrightarrow[]{b \times Id_{\mathbb{R}_{+}}} \xi \times \mathbb{R}_{+} \xrightarrow[]{\varGamma} \eta,\\ & \Phi_{0}: M^{n}_{0} \times \mathbb{R}_{+} \xrightarrow[]{f_{0} \times Id_{\mathbb{R}_{+}}} X^{n} \times \mathbb{R}_{+} \xrightarrow[]{\varGamma} F_{0},\\ & B_{0}: \nu_{M^{n}_{0}} \times \mathbb{R}_{+} \xrightarrow[]{b_{0} \times Id_{\mathbb{R}_{+}}} \xi_{0} \times \mathbb{R}_{+} \xrightarrow[]{\varGamma} \eta_{0}. \end{align*}

Their mapping cylinders $M(\Phi ,\, B)$ (respectively $M(\Phi _{0},\, B_{0})$) are normal spaces with boundaries $({M^{n}} \times \mathbb {R}_{+}) \amalg F_{0}$ (respectively $({M^{n}_{0}} \times \mathbb {R}_{+}) \amalg F_{0}$). Gluing them along $F_{0}$ yields the normal space

\[ N=M(F, B) \underset{F_{0}}{\cup} - M(F_{0}, B_{0}), \quad \partial N= {M^{n}} \times \mathbb{R}_{+}\ \underset{F_{0}}{\cup}\ {M^{n}_{0}} \times \mathbb{R}_{+}, \]

where the minus sign denotes the opposite orientation on $M(F_{0},\,B_{0})$.

This normal space $N$ can be decomposed into adic pieces to define an element in $H^{lf}_{n+2}(F_{0};\mathbb{\Omega}^{NSTOP})$, where $\mathbb{\Omega}^{NSTOP}$ is the semi-simplicially defined spectrum of adic normal spaces with manifold boundary (cf. Kühl, Macko and Mole [Reference Kühl, Macko and Mole12, Section 11] for the precise definition).

There is a similar spectrum $\mathbb{\Omega}^{NPD},$ where the boundaries are Poincaré duality spaces, and there exists an obvious map $\mathbb{\Omega}^{NSTOP} \to \mathbb{\Omega}^{NPD}.$ Moreover, there is a map of spectra $\mathbb{\Omega}^{NPD} \to \mathbb {L}^{+}$ (cf. Ranicki [Reference Ranicki19, p.287]), inducing isomorphisms in homology theory (cf. Hausmann and Vogel [Reference Hausmann and Vogel7], Levine [Reference Levitt13], Quinn [Reference Quinn18]). The composition $\mathbb{\Omega}^{NSTOP} \to \mathbb{\Omega}^{NPD} \to \mathbb {L}^{+}$ is called $sign^{\mathbb {L}}$ in Kühl, Macko and Mole [Reference Kühl, Macko and Mole12, p.232].

A word about notation: we shall denote the element represented by $M(\Phi ,\, B) \underset {F_{0}}{\cup } -M(\Phi _{0},\, B_{0})$ by

\[ \{f,b\}-\{f_{0},b{_0}\} \in H^{lf}_{n+2}(F_{0};\mathbb{\Omega}^{NSTOP}) \]

and its image under

\begin{align*} & H^{lf}_{n+2}(F_{0};\mathbb{\Omega}^{NSTOP}) \xrightarrow[]{\cong} H^{st}_{n+2}(F,X^{n},\{c_{0}\};\mathbb{\Omega}^{NSTOP})\\ & \xrightarrow[{\cong}]{\partial} H^{st}_{n+1}(X^{n};\mathbb{\Omega}^{NSTOP}) \xrightarrow[]{sign^{\mathbb{L}}} H^{st}_{n}(X^{n};\mathbb{L}^{+}) \end{align*}

will be denoted by $[f,\,b]-[f_{0},\,b_{0}].$

Finally, one can then show that the map $t:\mathcal {N}(X^{n}) \to H^{st}_{n} ( X^{n}; \mathbb {L}^{+})$ sending $(f,\,b)$ to $[f,\,b]-[f_{0},\,b_{0}],$ is well defined (cf. Hegenbarth and Repovš [Reference Hegenbarth and Repovš9, Theorem 5.1]).

3. Factorization of the map $t:\mathcal {N}(X^{n}) \to H^{st}_{n} ( X^{n}; \mathbb {L}^{+})$

This section is devoted to studying diagram (1.2).

I. First, one has to define the map $\mathcal {N}(X^{n}) \to \mathcal {N}(M^{n}_{0}).$ We shall keep the notation from § 2, so $(f_{0},\,b_{0})$ denotes the canonical surgery problem for an oriented closed generalized $n$-manifold $X^{n}$ with $f_{0}:M^{n}_{0} \to X^{n},\, \ b_{0}:\nu _{M^{n}_{0}} \to \xi _{0}.$

Let $(f,\,b)$ represent an element in $\mathcal {N}(X^{n})$, where $f:M^{n} \to X^{n},\, \ b:\nu _{M^{n}} \to \xi .$ We shall also write $\xi _{0},\, \xi :X^{n} \to BSTOP$ for the corresponding classifying maps. Their compositions with $\mathcal {J}: BSTOP \to BSG$ are homotopic.

Consider now the bundles $(f_{0})^{*}(\xi _{0})$ and $(f_{0})^{*}(\xi )$ over $M^{n}_{0}$. Observe that $(f_{0})^{*}(\xi _{0}) = \nu _{M^{n}_{0}}$ and that $(f_{0})^{*}(\xi )$ is fibre homotopy equivalent to $\nu _{M^{n}_{0}}.$ In other words, $(f_{0})^{*}(\xi )$ is a $TOP$-reduction of the Spivak fibration of the manifold $M^{n}_{0}$.

Therefore $(f_{0})^{*}(\xi )$ defines a surgery problem $f':{M'}^{n} \to M^{n}_{0},\, \ b':\nu _{{M'}^{n}} \to (f_{0})^{*}(\xi ),\,$ which we shall denote by $(f',\,b').$ These are well-known constructions (cf. Browder [Reference Browder1, Section II.4], Madsen and Milgram [Reference Madsen and Milgram14, Chapter 2], Wall [Reference Wall22, Chapter 10]).

Lemma 3.1 The composition of the normal maps

\[ {M'}^{n} \xrightarrow[]{f'} M^{n}_{0} \xrightarrow[]{f_{0}} X^{n} ,\quad \nu_{{M'}^{n}} \xrightarrow[]{b'} (f_{0})^{*}(\xi) \xrightarrow[]{\tilde{f}_{0}} \xi, \]

where $\tilde {f}_{0}$ is the obvious bundle map covering the map $f_{0},$ is normally bordant to $(f,\,b).$

Proof. For the proof, we have to describe $(f_{0},\,b_{0}),\,(f,\,b),$ and $(f',\,b')$ in more details. Suppose that $X^{n}$ is embedded into $S^{m},$ for some sufficiently large $m\ge n$, with a regular neighbourhood $W^{m}\subset S^{m}$ and a retraction $r:W^{m} \twoheadrightarrow X^{n}$. Thus, ${\left .{r}\right |_{\partial W^{m}}}: {\partial W^{m}} \twoheadrightarrow X^{n}$ is homotopy equivalent to the spherical fibration $\nu _{X^{n}}$, giving rise to $\beta : S^{m} \to W^{m}/\partial W^{m} \to T(\nu _{X^{n}}).$

The $TOP$-reductions $\xi _{0}$ and $\xi$ of $\nu _{X^{n}}$ then yield the following homotopy commutative diagram

(3.1)

Note that $h: T(\xi _{0}) \to T(\xi )$ is induced by a fibre homotopy equivalence $\dot {\xi }_{0} \sim \nu _{X^{n}} \sim \dot {\xi },$ where $\dot {\xi }_{0}$ (respectively $\dot {\xi }$) denotes the sphere bundles of $\xi _{0}$ (respectively $\xi$).

Denote the compositions with $\beta$ by $\alpha _{0}:S^{m} \to T(\xi _{0}),\, \ \alpha :S^{m} \to T(\xi ).$ They can be made transverse to $X^{n} \subset T(\xi _{0})$ (respectively $T(\xi )$) in order to obtain $\alpha ^{-1}_{0}(X^{n})=M^{n}_{0}$ (respectively $\alpha ^{-1}(X^{n})=M^{n}$), and $b_{0}$ (respectively $b$) are the obvious maps from their normal bundles in $S^{m}.$ Moreover, $\alpha _{0}$ (respectively $\alpha$) factor as $S^{m} \to T(\nu _{M^{n}_{0}}) \to T(\xi _{0})$ (respectively $S^{m} \to T(\nu _{M^{n}}) \to T(\xi )$) and we have the following homotopy commutative diagram

(3.2)

Note that $h: T(\xi _{0}) \to T(\xi )$ induces a homotopy equivalence $\bar {h}: T((f_{0})^{*}(\xi _{0})) \to T((f_{0})^{*}(\xi )).$ However, $(f_{0})^{*}(\xi _{0})=\nu _{M^{n}_{0}},$ so we get the following homotopy commutative diagram

(3.3)

Here, $\tilde {f}_{0}: (f_{0})^{*}(\xi ) \to \xi$ (respectively $\tilde {f}_{0}: (f_{0})^{*}(\xi _{0}) \to \xi _{0}$) are the obvious bundle maps over $f_{0}:M^{n}_{0} \to X^{n}$ (for simplicity we use the same symbol $\tilde {f}_{0}$ for both maps), and $T(\tilde {f}_{0})$ is the induced map between the Thom spaces, so $T(\tilde {f}_{0})^{-1}(X^{n})=M^{n}_{0},$ similarly for $T(\tilde {f}_{0}):T((f_{0})^{*}(\xi _{0})) \to T(\xi _{0}).$ Note that $h$ and $\bar {h}$ are not induced by bundle maps.

By making the composition

\[ S^{m} \xrightarrow[]{\alpha'} T(\nu_{M^{n}_{0}}) = T((f_{0})^{*}(\xi_{0})) \xrightarrow[]{\bar{h}} T((f_{0})^{*}(\xi)) \]

transverse to $M^{n}_{0},$ one obtains the surgery problem

\[ {M'}^{n}=(\bar{h}\circ {\alpha'})^{{-}1}(M^{n}_{0}) \to M^{n}_{0}, \quad b':\nu_{{M'}^{n}} \to (f_{0})^{*}(\xi). \]

Homotopy commutativity of diagram (3.3) then implies that

\[ {M'}^{n} \xrightarrow[]{f'} M^{n}_{0} \xrightarrow[]{f_{0}} X^{n}, \quad \nu_{{M'}^{n}} \xrightarrow[]{b'} (f_{0})^{*}(\xi) \xrightarrow[]{\tilde{f}_{0}} \xi \]

is normally bordant to $(f,\,b).$ To see this, observe that $(f,\,b)$ is obtained from the upper arrow $\alpha$, whereas the composition $(f_{0},\,b_{0}) \circ {(f',\,b')}$ is obtained from the composition of the arrows $\downarrow _{\longrightarrow }\uparrow$, that is $T({\tilde {f}}_{0}) \circ \bar {h} \circ {\alpha '}.$ Note that $T({\tilde {f}}_{0})$ produces $(f_{0},\,b_{0})$ and $\bar {h}\circ \alpha '$ gives $(f',\,b')$. This completes the proof of Lemma 3.1.

The association $(f,\,b) \to (f',\,b')$ defines a map $\mathcal {N}(X^{n}) \to \mathcal {N}(M^{n}_{0}).$ It depends on the fixed surgery problems $(f_{0},\,b_{0}),$ and $Id_{M^{n}_{0}} : M^{n}_{0} \xrightarrow []{\cong } M^{n}_{0},\, \ Id_{\nu _{M^{n}_{0}}}: \nu _{M^{n}_{0}} \xrightarrow []{\cong } \nu _{M^{n}_{0}}.$ We shall relate this map using the following identifications (cf. Kühl, Macko and Mole [Reference Kühl, Macko and Mole12, Chapter 14, in particular Section 14.23])

\[ \mathcal{N}(X^{n}) \to [X^{n}, G/TOP], \quad \mathcal{N}(M^{n}_{0}) \to [M^{n}_{0}, G/TOP]. \]

Given $f:{M}^{n} \to X^{n},\, \ b:\nu _{{M}^{n}} \to \xi ,\,$ we know that $\xi \oplus (-\xi _{0}): X^{n} \to BTOP$ classifies the Whitney sum of $\xi$ and $-\xi _{0}$. The composition with $\mathcal {J}:BTOP \to BSG$ is homotopic to the constant map, hence it yields a map $X^{n} \to G/TOP.$ This defines a bijection $\mathcal {N}(X^{n}) \to [X^{n},\,G/TOP],$ depending on $(f_{0},\,b_{0}).$

Let us denote the image of $(f,\,b)\in \mathcal {N}(X^{n})$ in $[X^{n},\,G/TOP]$ by $[\xi - \xi _{0}].$ Similarly, $\mathcal {N}(M^{n}_{0}) \to [M^{n}_{0},\,G/TOP]$ can be defined using $Id_{ M^{n}_{0}}: M^{n}_{0} \xrightarrow []{\cong } M^{n}_{0},\, \ Id_{\nu _{M^{n}_{0}}}: \nu _{M^{n}_{0}} \xrightarrow []{\cong } \nu _{M^{n}_{0}}.$ The construction above then implies the following corollary.

Corollary 3.3 The diagram

(3.4)

commutes. Moreover, $(f_{0})^{*}([\xi - \xi _{0}])= [(f_{0})^{*}(\xi ) - \nu _{M^{n}_{0}}].$

II. Next, we shall show how $(f',\,b')$ can be used to calculate $t(f,\,b)\in H^{st}_{n}(X^{n}; \mathbb {L}^{+}).$ By crossing $(f',\,b')$ with $\mathbb {R}_{+},$ one gets a normal map

\[ f'\times Id_{\mathbb{R}_{+}}:{M'}^{n}\times \mathbb{R}_{+} \to M^{n}_{0}\times \mathbb{R}_{+}, \quad b'\times Id_{\mathbb{R}_{+}}:\nu_{{M'}^{n}}\times \mathbb{R}_{+} \to (f_{0})^{*}(\xi)\times \mathbb{R}_{+}, \]

denoted by $(f',\,b')\times Id_{\mathbb {R}_{+}}.$ The mapping cylinder $M((f',\,b')\times Id_{\mathbb {R}_{+}})$ of the map $(f',\,b')\times Id_{\mathbb {R}_{+}}$ is a normal space with manifold boundary, hence it defines an element

\[ M((f',b')\times Id_{\mathbb{R}_{+}})\in H^{lf}_{n+2}(M^{n}_{0}\times \mathbb{R}_{+}; \mathbb{\Omega}^{NSTOP}). \]

Lemma 3.4 Let $\varGamma _{0}:M^{n}_{0}\times \mathbb {R}_{+} \to F_{0}$ be defined as the composition of the maps

\[ f_{0}\times Id_{\mathbb{R}_{+}}: M^{n}_{0}\times \mathbb{R}_{+} \to X^{n}\times \mathbb{R}_{+}, \quad \varGamma: X^{n}\times \mathbb{R}_{+} \to F_{0}. \]

Then $\varGamma _{0}$ induces a homomorphism

\[ (\varGamma_{0})_{*}: H^{lf}_{n+2}(M^{n}_{0}\times \mathbb{R}_{+}; \mathbb{\Omega}^{NSTOP}) \to H^{lf}_{n+2}(F_{0}; \mathbb{\Omega}^{NSTOP}), \]

such that

\[ (\varGamma_{0})_{*}([M((f',b')\times Id_{\mathbb{R}_{+}})])= \{f,b\}-\{f_{0},b_{0}\}. \]

Proof. The element $(\varGamma _{0})_{*}([M((f',\,b')\times Id_{\mathbb {R}_{+}})])$ is represented by the mapping cylinder

\[ ({M'}^{n}\times \mathbb{R}_{+})\times I \ \underset{f'\times Id_{\mathbb{R}_{+}}}{\cup} \ M^{n}_{0}\times \mathbb{R}_{+}, \]

but decomposed according to the dissection given by $\varGamma _{0}: M^{n}_{0}\times \mathbb {R}_{+} \to F_{0}.$ The element $\{f,\,b\}-\{f_{0},\,b_{0}\}$ is represented by

\[ ({M'}^{n}\times \mathbb{R}_{+})\times I\ \underset{\Phi}{\cup}\ F_{0} \ \underset{F_{0}}{{\cup}} \ -(M^{n}_{0}\times \mathbb{R}_{+})\times I\ \underset{\Phi_{0}}{\cup}\ F_{0}, \]

as described in § 2. By Lemma 3.1, it is equivalent to the mapping cylinder construction based on the composition of the normal maps $(f_{0},\,b_{0})\circ (f',\,b').$ It gives the following

\[ ({M'}^{n}\times \mathbb{R}_{+})\times I\ \underset{f'\times Id_{\mathbb{R}_{+}}}{\cup}\ M^{n}_{0}\times \mathbb{R}_{+} \cup (M^{n}_{0}\times \mathbb{R}_{+})\times I \underset{\varGamma_{0}}{\cup} \ F_{0} \ \cup{-}(M^{n}_{0}\times \mathbb{R}_{+})\times I\ \underset{\varGamma_{0}}{\cup}\ F_{0}, \]

(cf. Ferry [Reference Ferry4, Proposition 8.10] for the mapping cylinder calculations).

This is obviously bordant to

\[ ({M'}^{n}\times \mathbb{R}_{+})\times I\ \underset{f'\times Id_{\mathbb{R}_{+}}}{\cup}\ M^{n}_{0}\times \mathbb{R}_{+} \]

since

\[ (M^{n}_{0}\times \mathbb{R}_{+})\times I \underset{\varGamma_{0}}{\cup} \ F_{0} \ \cup{-}(M^{n}_{0}\times \mathbb{R}_{+})\times I\ \underset{\varGamma_{0}}{\cup}\ F_{0} \]

is $0$-bordant. This completes the proof of Lemma 3.4.

Now $(f',\,b')$ is a normal degree one map between manifolds, so it defines an element $[f',\,b']\in H_{n}(M^{n}_{0};\mathbb {L}^{+})$, namely its normal invariant.

Proof. The assertion follows from the following diagram

(3.5)

Note that the element $[M((f',\,b')\times Id_{\mathbb {R}_{+}})] \in H^{lf}_{n+2}(M^{n}_{0}\times \mathbb {R}_{+}; \mathbb{\Omega}^{NSTOP})$ maps to $[f',\,b']$ under the left vertical arrow of morphisms. The completion of $\varGamma _{0}$ then gives the map $\bar {\varGamma }_{0}: M^{n}_{0} \times [0,\,\infty ] \to F.$ This completes the proof of Corollary 

3.5

.

III. Summary: Let $X^{n}$ be an oriented closed generalized manifold of dimension $n\ge 5,$ and $f_{0}:M^{n}_{0} \to X^{n},\, \ b_{0}:\nu _{M^{n}_{0}}\to \xi _{0}$ a surgery problem according to a $BSTOP$-reduction of $\nu _{X^{n}}.$ Then the map $t:\mathcal {N}(X^{n}) \to H^{st}_{n} ( X^{n}; \mathbb {L}^{+}),$ defined in § 2, fits into the following commutative diagram

(3.6)

Here, $t_{0}$ sends a normal degree one map with target $M^{n}_{0}$ to its normal invariant. Moreover, under the identification of Corollary 3.3, diagram (3.6) can be redrawn as follows

(3.7)

4. Proof of Theorem 1.2

The essence of the proof was already given in § 1, by comparing diagrams (1.1) and (1.2). In this section, we present the details.

Let $\mathbb {L}^{\bullet }$ denote the symmetric $\mathbb {L}$-spectrum (cf. Ranicki [Reference Ranicki20, Chapter 13]). It is a ring spectrum and $\mathbb {L}^{+}$ is a $\mathbb {L}^{\bullet }$-module spectrum. Hence, the cup product constructions $H^{q}(Z,\,A; \mathbb {L}^{\bullet }) \times H^{p}(Z; \mathbb {L}^{+}) \to H^{p+q}(Z,\,A; \mathbb {L}^{+})$ are well defined.

Considering an oriented closed generalized $n$-manifold, embedded in $X^{n} \subset S^{m},$ for some $m \ge n,$ its Spivak fibration $\nu _{X^{n}}$ has a canonical orientation (cf. Ranicki [Reference Ranicki20, Chapter 16]), i.e. a Thom class

\[ \mathcal{U}_{\nu_{X^{n}}} \in H^{m-n}(E(\nu_{X^{n}}), \partial E(\nu_{X^{n}}); \mathbb{L}^{{\bullet}}) \cong \tilde{H}^{m-n}(T(\nu_{X^{n}}); \mathbb{L}^{{\bullet}}), \]

inducing the Thom isomorphism (here, $E(\nu _{X^{n}})$ is the associated disk fibration)

\[ H^{0}(X^{n}; \mathbb{L}^{+})= H^{0}(E(\nu_{X^{n}}); \mathbb{L}^{+}) \xrightarrow[]{\cup \ \mathcal{U}_{\nu_{X^{n}}}} \tilde{H}^{m-n}(T(\nu_{X^{n}}); \mathbb{L}^{+}). \]

Recall that canonical means that it is constructed via the canonical reduction $\xi _{0}$ of $\nu _{X^{n}}.$ Hence, the Thom class $\mathcal {U}_{\xi _{0}} \in \tilde {H}^{m-n}(T(\xi _{0}); \mathbb {L}^{\bullet }),\,$ corresponds to $\mathcal {U}_{\nu _{X^{n}}}$ under the homotopy equivalence between $T(\xi _{0})$ and $T(\nu _{X^{n}})$.

The existence of $\mathcal {U}_{\xi _{0}}$ is guaranteed (cf. Ranicki [Reference Ranicki20, Chapter 16]). Moreover, since $(f_{0})^{*}(\xi _{0}) \cong \nu _{M^{n}_{0}},$ it follows that $f_{0}:M^{n}_{0} \to X^{n},\, \ b_{0}:\nu _{M^{n}_{0}}\to \xi _{0}$ induces $T(b_{0}):T(\nu _{M^{n}_{0}}) \to T(\xi _{0}),$ so that under

\[ (T(b_{0}))^{*}:H^{m-n}(T(\xi_{0}); \mathbb{L}^{{\bullet}}) \to H^{m-n}(T(\nu_{M^{n}_{0}}); \mathbb{L}^{{\bullet}}), \]

$\mathcal {U}_{\xi _{0}}$ is mapped to $\mathcal {U}_{\nu _{M^{n}_{0}}}$, the Thom class of $\nu _{M^{n}_{0}}.$ This implies commutativity of the following diagram

(4.1)

The Thom isomorphisms are now composed with the $S$-duality isomorphisms:

\[ \tilde{H}^{m-n}(T(\nu_{M^{n}_{0}}); \mathbb{L}^{+}) \cong H^{st}_{n}(M^{n}_{0}; \mathbb{L}^{+}) \cong H_{n}(M^{n}_{0}; \mathbb{L}^{+}) \]

and

\[ \tilde{H}^{m-n}(T(\nu_{X^{n}}); \mathbb{L}^{+}) \cong H^{st}_{n}(X^{n}; \mathbb{L}^{+}). \]

For the generalized manifold $X^{n},$ this follows from Kahn, Kaminker and Schochet [Reference Kahn, Kaminker and Schochet10, Theorem B], which asserts that

\[ H^{m-n-1}(S^{m}\setminus X^{n};\mathbb{L}^{+} ) \cong H^{st}_{n}(X^{n}; \mathbb{L}^{+}). \]

Since for every $m \ge n,$

\[ H^{m-n-1}(S^{m};\mathbb{L}^{+})=L_{m-1}, \quad H^{m-n}(S^{m};\mathbb{L}^{+})=L_{m}, \]

where $L_{q}=\pi _{q}(G/TOP),$ the exact sequence of the pair $(S^{m},\,S^{m}\setminus X^{n})$ then implies that

\begin{align*} & H^{m-n-1}(S^{m}\setminus X^{n};\mathbb{L}^{+}) \cong \tilde{H}^{m-n}(S^{m},S^{m}\setminus X^{n};\mathbb{L}^{+})\\ & \quad \cong \tilde{H}^{m-n}(T(\nu_{X^{n}});\mathbb{L}^{+} ) \cong \tilde{H}^{m-n}(T(\xi_{0});\mathbb{L}^{+} ). \end{align*}

This also applies to $M^{n}_{0}.$

The proof of Kahn, Kaminker and Schochet [Reference Kahn, Kaminker and Schochet10, Theorem B] shows that the following diagram commutes

(4.2)

Briefly, this follows since the Spanier–Whitehead duality isomorphism comes from the slant product constructions, using the map

\[ X^{n}_{+} \wedge T(\xi_{0}) \cong X^{n}_{+} \wedge T(\nu_{X^{n}}) \to S^{m}, \]

i.e. it comes from the element in $H^{m}(X^{n}_{+} \wedge T(\xi _{0});\mathbb {L}^{+})$ which it defines. This construction is natural for the normal map $(f_{0},\,b_{0}).$ Since $X^{n}$ is not a complex, $T(\nu _{X^{n}})$ is replaced by a certain function space which leads to the Steenrod homology (cf. Kahn, Kaminker and Schochet [Reference Kahn, Kaminker and Schochet10, Section 4]).

Summary: The following diagram commutes

(4.3)

The composition of the upper row isomorphisms is known to be

\[ . \cap [M^{n}_{0}]_{\mathbb{L}^{{\bullet}}}: H^{0}(M^{n}_{0};\mathbb{L}^{+}) \xrightarrow[]{\cong} H_{n}(M^{n}_{0};\mathbb{L}^{+}), \]

where $[M^{n}_{0}]_{\mathbb {L}^{\bullet }} \in H_{n}(M^{n}_{0};\mathbb {L}^{\bullet })$ is the $\mathbb {L}^{\bullet }$-coefficient fundamental class of $M^{n}_{0}$ (cf. Ranicki [Reference Ranicki20, Proposition 18.3]). Finally, we can identify

\[ [M^{n}_{0},G/TOP]= H^{0}(M^{n}_{0};\mathbb{L}^{+}), \quad [X^{n},G/TOP]= H^{0}(X^{n};\mathbb{L}^{+} ), \]

according to the equivalence $G/TOP \xrightarrow []{\cong } \mathbb {L}^{+}$ (cf. Kirby and Siebenmann [Reference Kirby and Siebenmann11, Essay 5, Appendix C], Ranicki [Reference Ranicki20, Proposition 16.1]).

Combining this with Corollary 3.3 and diagram (3.6), we obtain the following diagram

(4.4)

Commutativity of the outer diagram (cf. diagram (4.3)) and each square imply that

\[ \mathcal{N}(X^{n}) \to \mathcal{N}(M^{n}_{0}) \xrightarrow[]{t_{0}} H_{n}(M^{n}_{0};\mathbb{L}^{+}) \xrightarrow[]{(f_{0})_{*}} H^{st}_{n}(X^{n};\mathbb{L}^{+}) \]

is an isomorphism, hence by diagram (3.6), this composition is $t.$ This completes the proof of Theorem 1.2.

5. Epilogue

We shall conclude this paper by a brief outlook for further studies, following a very interesting suggestion of the referee. In this paper, we have proved that there exists a bijective map $t:\mathcal {N}(X^{n}) \to H^{st}_{n} ( X^{n}; \mathbb {L}^{+})$ from normal degree one bordisms to the Steenrod homology of the spectrum $\mathbb {L}^{+}$.

The Steenrod homology is known to behave well on the category of compact metric spaces. In particular, if $q:X^{n}\to B$ is any morphism, then there exists a long exact sequence

(5.1)\begin{equation} \dots \to H^{st}_{n+1}(B;\mathbb{L}^{+}) \to H^{st}_{n+1}(B,X^{n};\mathbb{L}^{+}) \xrightarrow[]{\partial_{*}} H^{st}_{n}(X^{n};\mathbb{L}^{+}) \xrightarrow[]{q_{*}} H^{st}_{n}(B;\mathbb{L}^{+}) \to \dots \end{equation}

On the other hand, if $q:X^{n}\to B$ is a $UV^{1}$-map, then there is a controlled surgery sequence (cf. Bryant, Ferry, Mio and Weinberger [Reference Bryant, Ferry, Mio and Weinberger2], Mio [Reference Mio16], and Nicas [Reference Nicas17]),

(5.2)

Here, $\mathbb {L}$ denotes the $4$-periodic spectrum with $\mathbb {L}_{0}=\mathbb {Z} \times G/TOP$, $\sigma ^{c}$ is the controlled surgery obstruction map, and

(5.3)

is the controlled structure set. This controlled surgery sequence 5.2 makes sense if the controlled structure set

is nonempty (cf. Mio [Reference Mio16, Theorem 3.8]).

It is natural to ask if sequences (5.1) and (5.2) are related via the map $t:\mathcal {N}(X^{n}) \to H^{st}_{n} ( X^{n}; \mathbb {L}^{+})$. First, one notes that two spectra $\mathbb {L}^{+}$ and $\mathbb {L}$ are involved, where $\mathbb {L}^{+} \xrightarrow []{i} \mathbb {L}$ is considered as the covering spectrum over the Eilenberg–MacLane spectrum $K(\mathbb {Z},\,0)$, i.e. $\mathbb {L}^{+} \xrightarrow []{i} \mathbb {L}\to K(\mathbb {Z},\,0)$ is a fibration of spectra.

In order to compare sequences (5.1) and (5.2), we consider the composite map

\[ q_{*}\circ i_{*}: H^{st}_{n}(X^{n};\mathbb{L}^{+}) \xrightarrow[]{i_{*}} H^{st}_{n}(X^{n};\mathbb{L}) \xrightarrow[]{q_{*}} H^{st}_{n}(B;\mathbb{L}), \]

and obtain the following diagram

(5.4)

The first step would be to prove commutativity of diagram (

). However, this is not enough, since one also needs a map between ${H}^{st}_{n+1}(B,\, X^{n};\mathbb {L}^{+})$ and the set (5.3), compatible with $t:\mathcal {N}(X^{n}) \to H^{st}_{n} ( X^{n}; \mathbb {L}^{+})$. This can all be done if $X^{n}$ is a topological $n$-manifold (cf. Hegenbarth and Repovš [Reference Hegenbarth and Repovš8]). In the case when $X^{n}$ is a generalized $n$-manifold, this is still an unsolved problem.

For the second step, one is led to ”refining” the map $t:\mathcal {N}(X^{n}) \to H^{st}_{n} ( X^{n}; \mathbb {L}^{+})$ to a map

so that the following diagram is commutative

(5.5)

where ${\mathcal {S}^{c}}$ denotes the set (5.3).

Since $\dim X^{n}=n,$ we may assume that $\dim B\leq n.$ In this case, it follows from the Atiyah–Hirzebruch spectral sequence (which holds for the Steenrod homology, cf. Hegenbarth and Repovš [Reference Hegenbarth and Repovš9, p. 206]) that $H^{st}_{n+1}(B;\mathbb {L}^{+}) \xrightarrow []{i_{*}} H^{st}_{n+1}(B;\mathbb {L})$ is an isomorphism. In this case, the map

is bijective. However, the existence of such a map $\overline {t}$ is at present still a conjecture.