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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.

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Your definition is the same as Voevodsky's. Let us review his definition. Let $E\to X$ be a vector bundle of rank $r$ and $Th(E):= E/E-i(X)$. In the homotopy category of schemes $\mathbf{H}(X)$ we have an $\mathbb{A}^1$-equivalence $$ Th(E)\simeq \mathbb{P}(E\oplus 1)\mathbb{P}(E) $$ Voevodsky provides an elegant proof in "Reduced power operations..." above Proposition 4.3, you may also check Morel-Voevodsky 3.2, Proposition 2.17.

Therefore, we have split short exact sequencenes $$ 0\to H^{*,*}(Th(E))\xrightarrow{\varphi} H^{*,*}(\mathbb{P}(E\oplus1))\leftrightarrows H^{*,*}(\mathbb{P}(E))\to 0 $$ for any cohomology representable by spectra. We define the Thom class to be $\mathbf{t}(E)\in H^{2r,r}(Th(E))$ given by the formula $$ \varphi(\mathbf{t}(E))=\sum_{i=0}^r(-1)^{n-i}c_i(E)\cdot x^{n-i} \in H^{2r,r}(\mathbb{P}(E\oplus1)) \, , $$ where $x=c_1(\mathcal{O}_{\mathbb{P}(E\oplus 1)}(-1))$.

You will find in the literature some authors willing to emphasize when they consider the class in $H^{2r,r}(Th(E))$, for example calling it refined Thom class, and when they consider the class in $H^{2r,r}(\mathbb{P}(E\oplus 1)$.

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    $\begingroup$ thank you for your comments and answer. My supervisor said also Thom class can be denoted by linear combination of Chern class. $\endgroup$ – masa M Jul 12 '18 at 10:54

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