This is exercise I.2.4 from Basic Algebraic Geometry by Shafarevich.
Let $X$ be the curve defined by $y^2=x^2+x^3$ and $f:\mathbb{A}^1 \rightarrow X$ the map defined by $f(t)=(t^2-1, t(t^2-1))$. Prove that the corresponding homomorphism $f^*$ maps $k[X]$ isomorphically to the subring of the polynomial ring $k[t]$ consisting of polynomials $g(t)$ such that $g(1)=g(-1)$. (Assuming that char $k\neq 2$.
I can prove that $f^*$ is injective and its image is in this subring. To prove that it is onto, the only candidate for the pre-image of $g$ is $h:(x,y)\mapsto g(y/x)$. I don't know how to prove that $h$ defined as such is an element of $k[X]$, that is, it is the restriction of a polynomial in $k[x,y]$ to X.
I appreciate any help.
Let $$g(t)=a(t^2)+tb(t^2)$$ then if $g(1)=g(-1)$ we have
$$a(1)+b(1)=a(1)-b(1)$$
So $$b(1)=0$$ thus $b(t^2)=(t^2-1)c(t^2)$ and we have
$$g(t)=a(t^2)+t(t^2-1)c(t^2)=a(x+1)+yc(x+1).$$
So we have that
$$f^*(a(x+1)+yc(x+1))=g(t)$$