Equivalent definitions of the Euler class of an oriented vector bundle

Question

Let $E\to X$ be an oriented real $n$-plane bundle. Then the Euler class $e(E)\in H^n(X;\Bbb Z)$ can be defined, using Thom isomorphism (https://en.wikipedia.org/wiki/Euler_class#Formal_definition). I am curious about its properties, given here: https://en.wikipedia.org/wiki/Euler_class#Properties.

1. As Chern classes or Stiefel-Whitney classes, does the four properties (functoriality, Whitney sum formula, normalization, and orientation) uniquely characterize the Euler class?

2. It is also written that if $X$ is an oriented smooth $d$-manifold and $\sigma:X\to E$ is a smooth section that intersects the zero section transversally, then $e(E)$ is the Poincare dual of the class in $H_{d-r}(X;\Bbb Z)$ represented by the zero locus of $\sigma$. How can this be proved? Is there a reference of a proof for this statement?

Answer 1

The standard reference for characteristic classes is the book by Milnor and Stasheff "Characteristic Classes", and I recommend reading it.

Your second question is also answerd in these seminar notes ( Theorem 3.2) by Matthias Görg.

The answer to your first question is no. They don't make sure we just have $e(E)=0$ for all bundles.

Add a fifth axiom, saying something like $e(\gamma)[\mathbb{CP}^1] = -1 $ , where $\gamma \to \mathbb{CP}^1$ is the tautological bundle ( as an oriented $2$-plane bundle) and you can proove uniqueness, using a real splitting principle.

Edit: As pointed out in the comments, the uniqueness part is not working the way I thought. I posted this as a new question here.

Answer 2

(Edited after some discussion in the comments.)

As pointed out by Jonas, the four properties listed on Wikipedia do not characterise the Euler class as setting $e(E) = 0$ for all $E$ also satisfies those properties.

I just want to point out that the property which Wikipedia calls normalization is redundant: it follows from the Whitney sum formula and the orientation property. To see this, note that if $E \to X$ has a nowhere-zero section, then $E \cong E_0\oplus\varepsilon^1$ where $\operatorname{rank} E_0 = \operatorname{rank} E - 1$ and $\varepsilon^1$ is a trivial line bundle. If $E$ is orientable, then so is $E_0$, so by the Whitney sum formula $e(E) = e(E_0\oplus\varepsilon^1) = e(E_0)\cup e(\varepsilon^1)$. As $\varepsilon^1$ admits an orientation-reversing isomorphism, we have $e(\varepsilon^1) = e\left(\overline{\varepsilon^1}\right) = -e(\varepsilon^1)$, so $e(\varepsilon^1)$ is $2$-torsion, but $e(\varepsilon^1) \in H^1(X; \mathbb{Z})$ which is torsion-free, so $e(\varepsilon^1) = 0$ and hence $e(E) = e(E_0)\cup e(\varepsilon^1) = 0$.