Let $E$ be a vector bundle over a smooth scheme $X$. The Thom space of $E$ is $Th(E)=E/E-i(X)$ where $i\colon X \longrightarrow E$ is the zero section. This space is $\mathbb{A}^{1}$ isomorphic to $\mathbb{P}(E \oplus \mathscr{O})/\mathbb{P}(E)$

As in algebraic topology, the Thom class of a vector bundle $E$ (it is denoted $t_{E}$) is defined in motivic homotopy theory and satisfies the following property:

(Th) The Thom class gives the isomorphism $\tilde{H}^{*,*}(F_{･} \wedge X_{+}) \simeq \tilde{H}^{*+2\dim E,*+ \dim E}(F_{･} \wedge Th(E)) $

In Reduced power operations in motivic cohomology, V.Voevodsky defined the Thom class. But I don't understand it. So I think the definition of the Thom class is as follows.

The natural monomorphism between vector bundles $f \colon E \longrightarrow E\oplus \mathscr{O}$ gives a morphism $\mathbb{P}(f)\colon \mathbb{P}(E) \longrightarrow \mathbb{P}(E \oplus \mathscr{O})$ and $\mathbb{P}(f)^{*}(\mathscr{\sigma}_{E\oplus\mathscr{O}})=\sigma_{E}$ where $\sigma_{E}(resp. \sigma_{E\oplus \mathscr{O}}) =\mathscr{O}(-1) \in H^{2,1}(\mathbb{P}(E),\mathbb{Z})(resp.H^{2,1}(\mathbb{P}(E\oplus \mathscr{O}),\mathbb{Z}))$ if $\dim E=d $, then since $\{1,\sigma_{E},\cdots \sigma_{E}^{d-1}\}$ and $\{1,\sigma_{E\oplus \mathscr{O}},\cdots,\sigma_{E\oplus \mathscr{O}}^{d}\}$ are a basis of $H^{*,*}(\mathbb{P}(E),\mathbb{Z})$ and $H^{*,*}(\mathbb{P}(E\oplus \mathscr{O}),\mathbb{Z})$, $\mathbb{P}(f)^{*}$ is an epimorphism and the dimension of its Kernel 1. So the Thom class $t_{E}$ of $E$ is defined as the genereted element in its Kernel. As $Th(E) \simeq \mathbb{P}(E\oplus \mathscr{O})/\mathbb{P}(E)$, it belongs to $H^{*,*}(Th(E),\mathbb{Z})$.

Question. Is this definition right? If it isn't, I would like to know a correct definition of the Thom class.