We can find everywhere that $\phi :\mathbb A^1_K \to V(y^2-x^3)$ sending $t$ to $(t^2,t^3)$ defines a bijective map but not a morphism of affine varieties. To me it's not clear why it's even surjective. Here
Parametrization of the cuspidal cubic we say that we can parametrize the curve by considering a line passing through 0, and it's the argument I saw the most. I'm not used to parametrizations and I'm not sure to understand why this works, it looks arbitrary to me. How can we be sure that if we have $(u,v)\in V(y^2-x^3)$, then it's necessarily of this form ?
Also just to be sure, the reason of $K[x^2,x^3]\subsetneq K[x]$ (and so it's not an isomorphism of affine varieties) is because of the degree of elements in $K[x^2,x^3]$, which is either 0 or bigger than 2 is it correct ?
Thank you
"Each rational line $y=kx$ (where $k\neq0$) meets the variety (apart from the origin) at exactly one rational point. This shows that $\phi$ is an injective function.
Conversely, each rational point on the variety (leaving aside the origin) lies on a rational line $y=kx$ (where $k\neq0$). This shows that $\phi$ is a surjective function."
I'll let you formalise these statements yourself.
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$\phi$, as you defined it in the question, is indeed a morphism of affine varieties because $t^2,t^3\in K[t]$.
But for $K$ algebraically closed, $\phi$ is not an isomorphism of affine varieties, because indeed $t\notin K[t^2,t^3]$.