I read the following formal definition of Euler class see below. Two questions:
I suppose that $H^r(F, F \setminus F_0; \mathbf{Z})$ defines the relative cohomology --- why do we need this preciesly? Why do we need $(F, F \setminus F_0)$ to think about the orientation?
How should I think about the induced map from $ (X, \emptyset) \hookrightarrow (E, \emptyset) \hookrightarrow (E, E \setminus E_0), $ to $ H^r(E, E \setminus E_0; \mathbf{Z}) \to H^r(E; \mathbf{Z}) \to H^r(X; \mathbf{Z})$? And what is the role of the zero section $E_0$ here?
The Euler class ''e''(''E'') is an element of the integral [[cohomology]] group $$ H^r(X; \mathbf{Z}), $$ constructed as follows. An orientation of ''E'' amounts to a continuous choice of generator of the cohomology $$ H^r(F, F \setminus F_0; \mathbf{Z}) $$ of each fiber ''F'' [relative cohomology to the complement ''F''\''F''0 of its zero element ''F''0. From the Thom isomorphism, this induces an '''orientation class''' $$ u \in H^r(E, E \setminus E_0; \mathbf{Z}) $$ in the cohomology of ''E'' relative to the complement ''E''\''E''0 of the zero section ''E''0. The inclusions $$ (X, \emptyset) \hookrightarrow (E, \emptyset) \hookrightarrow (E, E \setminus E_0), $$ where ''X'' includes into ''E'' as the zero section, induce maps $$ H^r(E, E \setminus E_0; \mathbf{Z}) \to H^r(E; \mathbf{Z}) \to H^r(X; \mathbf{Z}).$$ The '''Euler class''' ''e''(''E'') is the image of ''u'' under the composition of these maps.
A choice of a local orientation of $\Bbb R^n$ at the origin is equivalent to choosing a vector space orientation of $T_0\Bbb R^n\simeq \Bbb R^n$ which is in turn equivalent to choosing a basis $(\mathbf{e}_1, \mathbf{e}_2, \cdots, \mathbf{e}_n)$ and remarking that every basis $\mathbf{b}$ of $\Bbb R^n$ is either positively or negatively orienting, depending upon if the linear isomorphism $(\Bbb R^n, \mathbf{e}) \to (\Bbb R^n, \mathbf{b})$ has positive or negative determinant, respectively. (If this happens, we will say $\mathbf{b}$ is compatible with $\mathbf{e}$)
This can be equated with the cohomological definition of local orientation: for any $n$-simplex $\triangle^n = \text{Nerve}(\{0, 1, 2, \cdots, n\}, <) \subset \Bbb R^n$, there is a natural local orientation on $T_p \triangle^n$ for any $p \in \text{int}\triangle^n$ given by a basis $\mathbf{v} = (\mathbf{v}_1, \cdots, \mathbf{v}_n)$ with $\mathbf{v}_k$ parallel to the edge realizations of $0 < k$ for $k = 1, \cdots, n$. For $M$ a smooth manifold, let's denote $\mathscr{C}^\bullet(M)$ and $\mathscr{C}_\bullet(M)$ to be the smooth singular (co)chain complexes, with chain groups freely generated by smooth maps $\triangle^\bullet \to M$, where $\triangle^n \subset \Bbb R^n$ is equipped with an induced smooth manifold-with-boundary structure. It's possible to show that the inclusion $\mathscr{C}_\bullet(M) \hookrightarrow C_\bullet(M)$ into the topological singular chain complex is a chain homotopy equivalence, so that the smooth (co)homology theory is naturally isomorphic to the singular (co)homology theory.
All of that being said, there is an (integral) $n$-cocycle in $\mathscr{C}^n(\Bbb R^n, \Bbb R^n\setminus \{0\})$ given by a functional $\psi : \mathscr{C}_n(\Bbb R^n,\Bbb R^n \setminus \{0\}) \to \mathbb{Z}$ which, for any smooth simplex $f : (\triangle^n, \partial \triangle^n) \to (\Bbb R^n, \Bbb R^n \setminus \{0\})$, takes value $\psi(f) = \pm 1$ depending on whether or not $df_*(\mathbf{v})$ is compatible with $\mathbf{e}$. By the previous discussion this gives rise to an element $[\psi]$ of $H^n(\Bbb R^n, \Bbb R^n\! \setminus \!\{0\})$. This is the cohomological local orientation class of $\Bbb R^n$ at the origin coming from the vector space orientation $[\mathbf{e}]$ of $T_0 \Bbb R^n \simeq \Bbb R^n$. In the end, it's a choice of sign assigned to if a given singular simplex is orientation-preserving or reversing on it's image.
Say $\pi : E \to M$ is a vector bundle of rank $k$ over a smooth manifold $M$. One can write down the cochain complex $\mathscr{C}_{vc}^\bullet(E)$ with the groups generated by cochains $\psi : \mathscr{C}_\bullet(E) \to \Bbb Z$ with vertical support, i.e., for every fiber $F_x \subset E$ there is a compact subset $0_x \subset K_x \subset F_x$ around the origin such that for any simplex $\sigma : \triangle^\bullet \to F_x \setminus K_x$, $\psi(\sigma) = 0$, i.e, $\psi$ is a compactly supported around the zero section $0_M \subset E$. Cohomology of this complex is the vertically supported cohomology $H_{vc}^\bullet(E)$ of the total space. This vector bundle will be called orientable if there is a choice of generator $g_x$ for every local homology group $H^n(F_x, F_x \setminus 0_x)$ in the fibers $F_x$, for $x \in M$, and these choices are coherent in the sense that if $x, y \in U \subset M$ are two points in a local trivialization of $\pi$, so that we have two isomorphisms $H^n(F_x, F_x \setminus 0_x) \to H^n(U \times F, U \times F \setminus U \times 0) \leftarrow H^n(F_y, F_y \setminus 0_x)$ by excision, then the images of $g_x$ and $g_y$ by the two arrows agree in the middle group.
If $(E, M, \pi)$ is an oriented bundle, then there is a pushforward map $\pi_*:H_{vc}^{n+k}(E) \to H^n(M)$ defined by "evaluating over orientation class of the fibers"; if $\psi$ is a vertically supported $(n+k)$-cocycle on $E$ and $\sigma : \triangle^n \to M$ a simplex in the base, then there is a natural extension of the simplex to the prism $\overline{\sigma} : \triangle^n \times \triangle^k \to M$ so that the $\triangle^k$-slices are orientation-preserving singular $k$-simplices inside the fibers of $\pi$ which contain the compact support of $\psi$ in the image (so the $\triangle^k$-slices are representatives of the generators of $H^n(F, F \setminus 0)$ that constitutes an orientation of $\pi$ over the simplex $\sigma$); explicitly $K_{\sigma(p)} \subset \overline{\sigma}(\{p\} \times \triangle^k) \subset F_{\sigma(p)}$ holds. The pushforward $\pi_* \psi$ is the cocyle in $\mathscr{C}^n(M)$ defined by $(\pi_*\psi)(\sigma) = \psi(\overline{\sigma})$, where $\overline{\sigma}$ has been appropriately triangulated. The cohomology class $[\pi_* \psi]$ is independent of the extension $\overline{\sigma}$ or the choice of triangulation.
According to the Thom isomorphism theorem, this is an isomorphism, and the inverse morphism $\Phi : H^n(M) \to H^{n+k}_{cv}(E)$ is given by $\Phi(\alpha) = \alpha \smile \mu$ for some fixed class $\mu = \Phi(1) \in H^k_{cv}(E)$, which follows from checking that $\pi_*$ satisfies naturality under cup product. This $\mu$ is known to be the Thom class of the vector bundle $(E, M, \pi)$. Since $\pi_* \Phi = \text{id}$, it must be that $\pi_* \Phi(1) = \pi_* \mu = 1$ $1 \in H^0(M)$, at the cochain level, is simply the constant functional $\mathbf{1}: \mathscr{C}_0(M) \to \Bbb Z$ which takes a $0$-simplex $p \in M$ and outputs the value $1$. But $\overline{\mathbf{1}}$ is simply an orientation-preserving simplex $\overline{\mathbf{1}} : \triangle^k \to F_p \subset E$ in the fiber over $p$ containing the compact support $K_p \subset F_p$ of $\mu$ in it's image. Since $\mathbf{1}(p) = 1$, by definition, $\mu(\overline{\mathbf{1}}) = 1$.
Therefore the more geometric definition of the Thom class $\mu \in H^k_{cv}(E)$ is a class which evaluates to (or integrates to, in the deRham setting) $1$ over $k$-simplices in $E$ along the fibers of $\pi$ containing the origin of the fiber in it's interior, and is compactly supported on a neighborhood of the zero section. In general there is an isomorphism $H^\bullet_{cv}(E) \to H^\bullet(E, E\setminus 0_M)$ by considering a vertically supported $k$-cocycle relative to the complement of the zero section. By excision it holds that $H^\bullet(E, E\setminus 0_M) \cong H^k(D(E), \partial D(E))$ where $D(E) \subset E$ is the disk subbundle of $E$ around the zero section $0_M \subset M$, it's boundary being the associated sphere bundle of $E$. This is isomorphic to $H^\bullet(\text{Thom}(E))$ where $\text{Thom}(E) \cong D(E)/\partial D(E)$ is the Thom space of $E$, alternatively obtained as $E^+/\infty_M$ from quotienting the fiberwise compactification of $E$ by the $\infty$-section.
An alternative description of $\mu \in H^k(E, E\setminus 0_M)$ would be as the Poincare dual of the zero section of $E$. This is because $\mu$ evaluates to $1$ over simplices in $E$ that lie transversely to $0_M$, which is precisely what Poincare duals do: If $Z^k \subset X^m$ is a closed submanifold the Poincare dual of $[Z] \in H_k(X)$ is defined as $\text{PD}[Z] \in H^{m-k}(X)$ such that $\text{PD}[Z] \frown [X] = [Z]$, where remember $\frown : H^\ell(X) \times H^n(X) \to H_{\ell-k}(X)$ is the cap product defined on chain level by setting $\psi^\ell \frown \sigma^n := \psi(\sigma[v_0, \cdots, v_\ell])\sigma[v_{\ell+1}, \cdots, v_n]$. This implies that in simplicial homology, if $[X]$ is written as a sum of $m$-simplices, then in $\text{PD}[Z] \frown [X]$ only the tangential-to-$Z$ $\ell$-slices of the $m$-simplices survive. This implies $\text{PD}[Z]$ must evaluate to $1$ along $(m-\ell)$-simplices in $X$ transverse to $Z \subset X$. But from the above discussion, that is precisely what $\mu$ does for $0_M \subset E$, therefore by uniqueness of Poincare duals, $\mu = \text{PD}[0_M]$
Finally, as mentioned in wikipedia, the Euler class $e(E)$ of the vector bundle $(E, M, \pi)$ is the image of $\mu$ under the sequence of maps $H^k(E, E \setminus 0_M) \to H^k(E) \to H^k(M)$ coming from the sequence of inclusions $(M, \emptyset) \to (E, \emptyset) \to (E, E\setminus 0_M)$. Suppose now that $s : M \to E$ is a general section of the vector bundle and consider the similar sequence of inclusions $$(M, \emptyset) \stackrel{s}{\hookrightarrow} (E, \emptyset) \hookrightarrow (E, E \setminus 0_M)$$ This is homotopic to the previous composition, because $s$ is homotopic to the zero section, simply by doing a fiberwise straightline homotopy. Therefore the image of $\mu$ by the induced map in cohomology is $e(E) \in H^k(M)$ as well. If $s$ is transverse to $0_M \subset E$, then $Z = s(M) \pitchfork 0_M$ in $s(M)$ be the zero set of the section $s$. By definition, $\mu$ evaluates to $1$ along $k$-simplices transverse to $0_M$ passing through points of $Z \subset s(M)$. But simplices in $M$ tangent to $s(M)$ at points of $Z$ which are also normal to $Z \subset M$, or more explicitly, simplices along fibers of the normal bundle $\nu_M(Z)$, are transverse to $0_M$ by hypothesis ($s \pitchfork 0_M$), so after pulling back $\mu$ by $s : M \to s(M)$, we obtain that $e(M) \in H^k(M)$ evaluates to $1$ along simplices normal to $Z \subset M$. Therefore $e(M) = \text{PD}(Z)$.
Euler class of a vector bundle is simply the Poincare dual of the zero locus of a generic section. $\blacksquare$