Injective map between two schemes Assuem we have a finite surjective map between two irreducible, separated schemes, $f:X \rightarrow Y$, and for a dense open $U \subset Y$ and for any $y \in U$, $|X_y| =1$, then can we say $f$ is injective everywhere?
We can also assume that $Y$ is normal.
 A: With some additional hypotheses this follows from Zariski's Main theorem.

If $f \colon X \to Y$ is a birational projective morphism between noetherian integral schemes, then the inverse image of every normal point of $Y$ is connected. [Hartshorne (1977, Corollary III.11.4)]

See also: https://en.wikipedia.org/wiki/Zariski%27s_main_theorem

If you make the assumption that $X$ and $Y$ are noetherian and integral, I guess you will be there.
A: The answer is "no" without some further hypothesis on $X$ and $Y$.  For instance, let $Y$ be $\text{Spec}(\mathbb{Z}_{5\mathbb{Z}})$, the localization of $\mathbb{Z}$ at the prime ideal $5\mathbb{Z}$, let $X$ be $\text{Spec}(\mathbb{Z}_{5\mathbb{Z}}[x]/\langle x^2 + 1 \rangle)$, and let $U$ be the distinguished open where $5$ is invertible.  
However, for finite type schemes over a field, this does follow from Zariski's Main Theorem.
Edit. I did not see jmc's answer below when I posted this answer.  This answer is basically the same as jmc's answer.  It is important to note, even for finite type schemes over a field, being injective on points does not necessarily imply the morphism is birational, e.g., Frobenius morphisms.
A: The situation is much worse; you don't need pathological examples. Just take any blow-up: the blow-down map is surjective and injective almost everywhere. I think your claim is true if both $X$ and $Y$ are projective curves.
Added in proof:
sorry, I meant by "finite" generically finite-to-one. Otherwise, "A variety X over a field is normal if and only if every finite birational morphism from any variety Y to X is an isomorphism", see Normal scheme
