Let $G$ be a group acting on a variety $X$ such that every $g\in G$ defines a morphism $\phi_g:X\rightarrow X$ given by $\phi_g(x)=g\cdot x$. If $X=\operatorname{Spec}(R)$ is affine then $\phi_g$ comes from a map $\phi_g^*:R\rightarrow R$. We define the induced action of $G$ on $R$ by $$g\cdot f=\phi_{g^{-1}}^{*}(f)$$ for $f\in R$
In other words, $(g\cdot f)(x)=f(g^{-1}\cdot x)$ for all $x\in X$
I have two questions -
Why do we need $g^{-1}$ to ensure that $G$ induces an action on $R$? We have - $$(g_1 g_2)\cdot f=\phi^*_{(g_1g_2)^{-1}}(f)=\phi^*_{g_2^{-1}g_1^{-1}}(f)=(\phi^*_{g_2^{-1}}\phi^*_{g_1^{-1}})(f)$$ At this point I am confused as $$(\phi^*_{g_2^{-1}}\phi^*_{g_1^{-1}})(f)=\phi^*_{g_2^{-1}}(\phi^*_{g_1^{-1}}(f))=\phi^*_{g_2^{-1}}(g_1\cdot f)=g_2\cdot(g_1\cdot f)$$ which is not what I want to show that it is an action. So what am I doing wrong?
Can someone explain to be how to show the "in other words" part?
Thank you.
(1) Your error is that $\phi^*_{g_2^{-1}g_1^{-1}}$ is $\phi^*_{g_1^{-1}}\phi^*_{g_2^{-1}}$, not $\phi^*_{g_2^{-1}}\phi^*_{g_1^{-1}}$. We have $\phi_{gh}=\phi_g\phi_h$, but passing from $\phi$ to $\phi^*$ is a contravariant functor, so it reverses the order of composition.
(2) Suppose $\phi:X\to X$ is a morphism and $x$ is a $k$-point of $X$. If $f\in R$, write $f(x)$ for the image of $f$ under the homomorphism $R\to k$ corresponding to $x$. Equivalently, $f(x)=i_x^*(f)$, where $i_x:\operatorname{Spec} k\to X$ is the map corresponding to $x$. Then $i_{\phi(x)}=\phi i_x$, so $$f(\phi(x))=i_{\phi(x)}^*(f)=i_x^*\phi^*f=(\phi^*f)(x).$$
In particular, in the case $\phi=\phi_{g^{-1}}$, the left-hand side is $f(g^{-1}\cdot x)$ and the right-hand side is $(g\cdot f)(x)$.