# Morphism which is a bijection on closed points

My question might be very naive.

Let $$X$$ and $$Y$$ be reduced schemes of finite type over an algebraically closed field $$k$$ and $$f:X\longrightarrow Y$$ a morphism of $$k$$-schemes. Let us assume that $$Y$$ is normal, connected and that $$f$$ induces a bijection on closed points. Can we conclude from these assumptions that $$f$$ is a local isomorphism?

If I assume that $$X$$ is also connected, may I conclude that $$f$$ is an isomorphism?

• No. Let $Y= \mathbb{A}^1_k$, and $X$ the disjoint union of $\mathbb{A}^1_k-\{0\}$ and $\{0\}$ with $f$ the obvious map. – Donu Arapura Apr 18 '19 at 22:05
• Ok, I just edited my question. – Gaussian Apr 18 '19 at 22:09

If $$k$$ is algebraically closed of non-zero characteristic $$p$$, then bijectivity on closed points is not enough (take the morphism $$\mathrm{Spec}\,k[x]\rightarrow \mathrm{Spec}\,k[x]$$ corresponding to $$f(T)\rightarrow f(T^p)$$).
If $$k$$ is algebraically closed of characteristic 0, then bijectivity on closed points is enough (if your target is normal, that is).
The reason why characteristics 0 and $$p$$ apparently behave differently is that only in $$\mathrm{char}\,p$$ you have this funny thing called purely inseparable field extension.