It's widely known that the affine variety $\mathcal{V}(X^3 - Y^2) \subset \mathbb{C}^2$, which graphs to a cubic curve with one cusp at the origin, is not isomorphic to $\mathbb{C}$, which has no cusps. There's an involved algebraic answer proving why the $\mathbb{C}$ and $\mathcal{V}(X^3 - Y^2)$ are not isomorphic here: Two affine varieties are not isomorphic
I'm wondering if anyone could provide a more geometric proof.
Can we safely conclude that if two curves embedded in the complex plane are not homeomorphic, then the varieties they represent are also not isomorphic? This is the definition I currently have: $X$ and $Y$ are isomorphic iff $\exists$ polynomial maps $g: X \rightarrow Y$ and $h: Y \rightarrow X$ st $h \circ g = id_X$ and $g \circ h = id_Y$. The definition looks suspiciously similar the definition for homotopy equivalence, but I'm having a hard time seeing the concrete connection. Could anyone explain it to me (if it exists)? Note: As noted in the comments, I can't use this fact (if true; it's still not entirely clear to me why it's the case) to prove this specific question.
Thanks in advance!
Generally speaking, you can talk about whether two objects are isomorphic in any category - for example, groups, rings, vector spaces, and here, affine varieties - and the definition is exactly the one you've written in all of those cases. For homotopy equivalences the relevant category is called the homotopy category of spaces.
And generally speaking, the way you show two objects are not isomorphic is by finding some property (invariant under isomorphism) shared by one object but not another. A basic property of affine varieties is the dimensions of Zariski tangent spaces at each point: you can compute that the dimension of the Zariski tangent space at each point of the affine line $\mathbb{A}^1$ is $1$, but the dimension of the Zariski tangent space of $V(X^3 - Y^2)$ at the cusp $(0, 0)$ is $2$. This reflects the fact that the cusp is singular.