A morphism of complex affine schemes which is bijective on complex points but is not an isomorphism. I am trying to find some such examples:
A morphism of complex affine schemes which is bijective on complex points but is not an isomorphism.
May I please ask for some examples, if possible. I would like to know how to construct such an example and how to think about it.
Thanks in advance!
 A: A bijection on complex points, but not a homeomorphism, as it is the case for the canonical bijection $\mathrm{Spec} \mathbb{C}\bigsqcup \mathbb{A}^1_\mathbb{C}\to\mathbb{P}^1_\mathbb{C}$. More generically, you can take an irreducible complex scheme $X$, an open subscheme $U\subset X$ with a complementary closed subscheme $V$, and consider the canonical morphism $V\bigsqcup U\to X$. 
A universal homeomorphism, but not an isomorphism, as it is the case for the normalisation of the cuspidal curve, mentioned in Mohan's comment. Also, the closed immersion of the maximal reduced closed subscheme $X_{\mathrm{red}}\to X$, for every non-reduced complex affine scheme $X$ provides such an example. The discussion on universal homeomorphisms here and here might also be useful in constructing such examples.
A: Here's a completely different flavor of counterexample.
Let $X = \operatorname{Spec}(\mathbb{C}(x))$. Then $X$ is an affine scheme over the complexes, but $X(\mathbb{C}) = \varnothing$ — it has no points at all with $\mathbb{C}$ as the residue field!
With the basic idea, it's easy to see how to construct lots of counterexamples.
A: Here's a more ad-hoc example: consider the morphism $f:\operatorname{Spec}(B) \to  \operatorname{Spec}(A)$ where:
$$ A = \mathbb{C}[x, x^{-1}] $$
$$ B = \mathbb{C}[y, y^{-1}] $$
and the morphism is induced by $x \to y^2$.
Geometrically, $\operatorname{Spec}(A)$ is the complex line with the origin removed, and $\operatorname{Spec}(B)$ is the double cover corresponding to the graph of the multi-valued function $\sqrt{x}$.
On maximal ideals, $f$ is precisely a 2-1 function. Let $S$ be any multiplicative subset of $B$ generated by the generators of exactly one ideal per pair; e.g. you can let $S$ be the set of all functions whose roots all have complex argument in $[0, \pi)$.
Then if we form the localization of $B$ by inverting the elements of $S$, the restriction $\operatorname{Spec}(S^{-1} B) \to \operatorname{Spec}(A)$ is bijective on points, and bijective on complex points.
