[proofplan]
We identify the Poincaré dual of $[N]$ by passing through the tubular neighbourhood theorem. The normal bundle $\nu_{N \subseteq M}$ embeds as an open neighbourhood $U$ of $N$ in $M$, and the fundamental class $[N]$ corresponds via excision and Poincaré duality to the Thom class $\mathcal{E}_{N \subseteq M}$ of the normal bundle. Extending the Thom class by zero to all of $M$ produces a class in $H^{d-n}(M; R)$ that equals $D_M^{-1}([N])$, verified by checking the cap product formula fiberwise.
[/proofplan]
[step:Embed the normal bundle as a tubular neighbourhood]
Let $M$ be a $d$-dimensional $R$-oriented manifold and $N \subset M$ a closed $n$-dimensional $R$-oriented submanifold. The normal bundle $\nu_{N \subseteq M}$ has fiber dimension $d - n$. By the tubular neighbourhood theorem, there exists an open neighbourhood $U$ of $N$ in $M$ and a diffeomorphism
\begin{align*}
\phi: \nu_{N \subseteq M} \xrightarrow{\;\sim\;} U
\end{align*}
that restricts to the identity on the zero section $N \hookrightarrow \nu_{N \subseteq M}$. This identifies $U$ with the total space of $\nu_{N \subseteq M}$, and the complement $U \setminus N$ corresponds to the deleted bundle $\nu_{N \subseteq M}^\#$.
[/step]
[step:Identify the Thom class via local cohomology and excision]
The Thom class $\mathcal{E}_{N \subseteq M} \in H^{d-n}(\nu_{N \subseteq M}, \nu_{N \subseteq M}^\#; R)$ is the unique class provided by the [Thom Isomorphism Theorem](/theorems/2283) whose restriction to each fiber $((\nu_{N \subseteq M})_x, (\nu_{N \subseteq M})_x^\#) \cong (\mathbb{R}^{d-n}, \mathbb{R}^{d-n} \setminus \{0\})$ is the orientation class $\varepsilon_x$.
Under the tubular neighbourhood identification $\phi: (\nu_{N \subseteq M}, \nu_{N \subseteq M}^\#) \cong (U, U \setminus N)$, the Thom class corresponds to a class $\phi^{-1*}\mathcal{E}_{N \subseteq M} \in H^{d-n}(U, U \setminus N; R)$.
[/step]
[step:Extend the Thom class by zero to obtain a global cohomology class]
Consider the long exact sequence of the pair $(M, M \setminus N)$. By excision (since $M \setminus U$ is a closed subset of $M \setminus N$), the inclusion $(U, U \setminus N) \hookrightarrow (M, M \setminus N)$ induces an isomorphism
\begin{align*}
H^{d-n}(M, M \setminus N; R) \xrightarrow{\;\sim\;} H^{d-n}(U, U \setminus N; R).
\end{align*}
Under this isomorphism, the Thom class defines a class $\tau \in H^{d-n}(M, M \setminus N; R)$. The natural map $H^{d-n}(M, M \setminus N; R) \to H^{d-n}(M; R)$ (induced by the inclusion $(M, \varnothing) \hookrightarrow (M, M \setminus N)$) sends $\tau$ to a class
\begin{align*}
\bar{\tau} \in H^{d-n}(M; R),
\end{align*}
which is the "extension by zero" of the Thom class to all of $M$. At the cochain level, a relative cocycle representing $\tau$ vanishes on chains supported in $M \setminus N$, so its image in $H^{d-n}(M; R)$ has support concentrated near $N$.
[/step]
[step:Verify that $\bar{\tau} = D_M^{-1}([N])$ via the cap product]
We must show that $D_M(\bar{\tau}) = [M] \frown \bar{\tau} = [N] \in H_n(M; R)$.
The inclusion $j: N \hookrightarrow M$ induces $j_*: H_n(N; R) \to H_n(M; R)$, and $[N]$ in $H_n(M; R)$ means $j_*[N]$. By [Poincaré Duality](/theorems/2291) on $M$, the map $D_M: H^{d-n}(M; R) \to H_n(M; R)$ is an isomorphism, so it suffices to check that $[M] \frown \bar{\tau} = j_*[N]$.
This verification is fiberwise. At each point $x \in N$, the local fundamental class $\mu_x \in H_d(M, M \setminus \{x\}; R)$ decomposes (via the product structure of the tubular neighbourhood) as a cross product of the local fundamental class of $N$ at $x$ with the local fundamental class of the fiber $\nu_x$. Capping $\mu_x$ with the restriction of $\tau$ to the fiber gives the local fundamental class of $N$ at $x$. Since $[M] \frown \bar{\tau}$ is the unique class in $H_n(M; R)$ that restricts to the local fundamental class of $N$ at every $x \in N$, and since $j_*[N]$ has the same local property by the [Fundamental Class Theorem](/theorems/2290) applied to $N$, we conclude $[M] \frown \bar{\tau} = j_*[N]$.
Therefore $D_M(\bar{\tau}) = [N]$, i.e., $D_M^{-1}([N]) = \bar{\tau}$, which is the extension by zero of the Thom class $\mathcal{E}_{N \subseteq M}$.
[guided]
The argument has two layers: a formal reduction using excision and the tubular neighbourhood theorem, and a local computation using the product structure.
The formal layer works as follows. The tubular neighbourhood theorem identifies a neighbourhood of $N$ in $M$ with the total space of the normal bundle. The Thom class lives in the relative cohomology $H^{d-n}(\nu, \nu^\#; R)$, which excision identifies with $H^{d-n}(M, M \setminus N; R)$. Forgetting the relative part (i.e., mapping to $H^{d-n}(M; R)$) produces the "extension by zero" $\bar{\tau}$.
The local computation verifies that capping $[M]$ with $\bar{\tau}$ recovers $[N]$. Why should this be true? Locally near a point $x \in N$, the manifold $M$ looks like $\mathbb{R}^n \times \mathbb{R}^{d-n}$ (the product of a piece of $N$ with a normal fiber), and $\bar{\tau}$ restricts to the generator of $H^{d-n}(\mathbb{R}^{d-n}, \mathbb{R}^{d-n} \setminus \{0\}; R)$ in the fiber direction. Capping the product fundamental class $[M]_{\text{local}} = [\mathbb{R}^n] \times [\mathbb{R}^{d-n}]$ with this fiber class "integrates out" the normal direction, leaving the tangential class $[\mathbb{R}^n]$, which is precisely the local fundamental class of $N$.
Since both $[M] \frown \bar{\tau}$ and $j_*[N]$ restrict to the same local class at every point of $N$, they must be equal by the uniqueness assertion in the [Fundamental Class Theorem](/theorems/2290).
[/guided]
[/step]
