How to show that $(\delta'f)(e)=\delta(\phi^* f)(e)$?

Question

I am reading linear algebraic groups.

I have a question in line 4 of the third paragraph of Page 66. How to show that $(\delta'f)(e)=\delta(\phi^* f)(e)$ for all $f\in K[G]$? Here $G$ is an algebraic group, $K$ is an algebraic closed field, $\delta' = \operatorname{Ad} x(\delta) = d(\operatorname{Int} x(\delta))$, $\phi=\operatorname{Int} x$, $\operatorname{Int} x(y) = xyx^{-1}$, $x, y \in G$. Thank you very much.

I think that $\delta'f = d(\operatorname{Int} x(\delta))f$.

Answer 1

Let $\varphi: X \to Y$ be a morphism of varieties. Then we have a map $\varphi^*: k[Y] \to k[X]$ defined by $f \mapsto f \circ \varphi$. The tangent map at some $x \in X$ is given by $d\varphi_x: \mathcal{T}_x(X) \to \mathcal{T}_{\varphi(x)}(Y)$, $\delta \mapsto \delta \circ \varphi^*$. Therefore $$ ((d(Int x))(\delta))(f) = (\delta (Int x)^*)(f) = \delta(f\circ Int x).$$