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You have omitted a crucial part of the sentence:

... in the sense of §16.

The point is small objects admit canonical basepoints via the augmentation of the Weil algebras defining them. See this MSE questionthis MSE question for the definition of $\operatorname{Spec}_R(\pi)$ for the augmentation $\pi:W\rightarrow R$. The author's definition only asks for pullbacks involving these canonical base points.

For the dual numbers, this notion does pick zero as a basepoint, since dual numbers can be identified with $R^2$ with-dual-number-multiplication, making the projection of $\bar x=(0,x)$ clearly zero.

You have omitted a crucial part of the sentence:

... in the sense of §16.

The point is small objects admit canonical basepoints via the augmentation of the Weil algebras defining them. See this MSE question for the definition of $\operatorname{Spec}_R(\pi)$ for the augmentation $\pi:W\rightarrow R$. The author's definition only asks for pullbacks involving these canonical base points.

For the dual numbers, this notion does pick zero as a basepoint, since dual numbers can be identified with $R^2$ with-dual-number-multiplication, making the projection of $\bar x=(0,x)$ clearly zero.

You have omitted a crucial part of the sentence:

... in the sense of §16.

The point is small objects admit canonical basepoints via the augmentation of the Weil algebras defining them. See this MSE question for the definition of $\operatorname{Spec}_R(\pi)$ for the augmentation $\pi:W\rightarrow R$. The author's definition only asks for pullbacks involving these canonical base points.

For the dual numbers, this notion does pick zero as a basepoint, since dual numbers can be identified with $R^2$ with-dual-number-multiplication, making the projection of $\bar x=(0,x)$ clearly zero.

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You have omitted a crucial part of the sentence:

... in the sense of §16.

The point is small objects admit canonical basepoints via the augmentation of the Weil algebras defining them. See this MSE question for the definition of $\operatorname{Spec}_R(\pi)$ for the augmentation $\pi:W\rightarrow R$. The author's definition only asks for pullbacks involving these canonical base points.

For the dual numbers, this notion does pick zero as a basepoint, since dual numbers can bybe identified with $R^2$ with-dual-number-multiplication, making the projection of $\bar x=(0,x)$ clearly zero.

You have omitted a crucial part of the sentence:

... in the sense of §16.

The point is small objects admit canonical basepoints via the augmentation of the Weil algebras defining them. See this MSE question for the definition of $\operatorname{Spec}_R(\pi)$ for the augmentation $\pi:W\rightarrow R$. The author's definition only asks for pullbacks involving these canonical base points.

For the dual numbers, this notion does pick zero as a basepoint, since dual numbers can by identified with $R^2$ with-dual-number-multiplication, making the projection of $\bar x=(0,x)$ clearly zero.

You have omitted a crucial part of the sentence:

... in the sense of §16.

The point is small objects admit canonical basepoints via the augmentation of the Weil algebras defining them. See this MSE question for the definition of $\operatorname{Spec}_R(\pi)$ for the augmentation $\pi:W\rightarrow R$. The author's definition only asks for pullbacks involving these canonical base points.

For the dual numbers, this notion does pick zero as a basepoint, since dual numbers can be identified with $R^2$ with-dual-number-multiplication, making the projection of $\bar x=(0,x)$ clearly zero.

Source Link
Arrow
  • 10.5k
  • 1
  • 27
  • 71

You have omitted a crucial part of the sentence:

... in the sense of §16.

The point is small objects admit canonical basepoints via the augmentation of the Weil algebras defining them. See this MSE question for the definition of $\operatorname{Spec}_R(\pi)$ for the augmentation $\pi:W\rightarrow R$. The author's definition only asks for pullbacks involving these canonical base points.

For the dual numbers, this notion does pick zero as a basepoint, since dual numbers can by identified with $R^2$ with-dual-number-multiplication, making the projection of $\bar x=(0,x)$ clearly zero.