An exercise about pullback morphism between rings of regular functions. Fix a field $K$ of characteristic zero. We are given the affine variety $X=V(y^2-x^2-x^3)\subset\mathbb{A}^2$ and the morphism $f:\mathbb{A}^1\rightarrow X$ given by $t\mapsto ((t^2-1),t(t^2-1))$ and we are asked to show that the pullback $f*:\mathscr{O}(X)\rightarrow K[t]$ maps isomoprhically $\mathscr{O}(X)$ in the subring $S$ of $K[t]$ made by $\\{g(t)\in K[t]\mid g(1)=g(-1)\\}$. I know little about how to deal with morphisms between algebras of which we know the presentation, but I think the image will be generated by the images of the generator, modulo the image of the relator, hence it will have generators $\\{(t^2-1),t(t^2-1)\\}$ and the relation $(t-1)^3(t+1)^4=0$, the fact is I do not know how to show, if this is the true image, that it coincides with $S$.

Since $X$ is affine, we have $\mathscr{O}(X) = K[x,y]/(y^2-x^2-x^3)$. The map $f^\ast \colon K[x,y]/(y^2-x^2-x^3) \rightarrow K[t] $ is defined by $f^\ast(x) = t^2-1$ and $f^\ast(y) = t(t^2-1)$. Hence for any $p(x,y) \in K[x,y]/(y^2-x^2-x^3)$ we have $$f^\ast p(t) = p\left(t^2-1, t(t^2-1) \right) = p(0,0) + (t^2-1)q(t)$$ In particular, $f^\ast p(t) \in S$ and we will show that $f^\ast\colon \mathscr{O}(X)\rightarrow S$ is an isomorphism.

By the first isomorphism theorem we only have to prove that the image of $f^\ast$ is $S$

Let $p(t) \in S$. By the division algorithm $p(t) = p_1(t)(t^2-1) + b_0$. As $p_1(t) = p_2(t)(t^2-1) +a_1t+b_1$ we have $$ p(t) = p_2(t)(t^2-1)^2 +(a_1t+b_1)(t^2-1) +b_0 $$ Iterating this process we show that $p(t) = q\left(t^2-1, t(t^2-1) \right)$ hence $p(t)$ is the image of (the class of) $q(x,y)$.