# On morphisms on varieties: 1-1 and projective implies iso?

For my thesis i have to learn about double covers of surfaces. All varieties in this question are projective over $\mathbb{C}$ if necessary.

Let $\phi: V \rightarrow W$ be a double cover of surfaces (by definition a surjective map, finite of degree 2), with ramification locus $\cup E_i$ on $W$ (where the curves $E_i$ are disjoint and their sum is a divisor divisible by two of course).

Now, using that $\phi$ is projective, hence universally closed, i can show that the preimage of an $E_i$ is a curve $F_i$ that is irreducible as well. Hence $\phi|_{F_i}: F_i \rightarrow E_i$ is a finite map of degree 1 between irreducible projective curves.

Now my question is: How do we show that $\phi|_{F_i}$ is an isomorphism?

I am giving the background because i expect that in general, finite 1-1 surjective projective maps need not be isomorphisms. When answering, could you say something about this as well? It would be really nice if it holds but the assumptions seem so weak...

Joachim

Edit:

• In fact, i do need to show that $\phi|_{F_i}$ is finite of degree one. Can someone help me out here?

• I am working on the general case now, and i think i showed that a finite 1-1 surjective projective map gives an open cover with isomorphic opens of both varieties. Can we use projectiveness to immediately conclude isomorphism of projective varieties?

I'm sorry i made a bit of a mess of this question, should have given it some more thought before posting it. Since people might have read it already i think its better to keep it in the current order.

P.S. I think i'm neglecting the proper use of preimage scheme-theoretically, which i think should include the double ramification in the structure sheaf. But i'm happy to take it the naive way. Any comments on this are appreciated but not necessary.

-
A finite map of degree $1$ between irreducible projective curves is an isomorphism: this is because such a map induces an isomorphism of the field of rational functions on the two curves, and this in turn gives an inverse rational map; but a rational map between two irreducible projective curves is automatically a morphism. – Zhen Lin Sep 23 '12 at 16:40
@ZhenLin wow thats nice! And something i was supposed to know already.. But it doesnt use the covering: so when we replace surface by variety of degree $n$, and curve by a codimension 1 subvariety does the same hold (with a different argument)? – Joachim Sep 23 '12 at 16:56
I don't know whether the map you have defined is really a finite map of degree $1$ – I'm just pointing out what you can do if you know that it is! – Zhen Lin Sep 23 '12 at 16:57
Right, i missed that important point. I will make an edit, thanks! – Joachim Sep 23 '12 at 17:14
Dear @Zhen Lin, your first comment is only correct for smooth curves : the normalization of a cusp is not an isomorphism and its inverse is a rational map which is not a morphism. – Georges Elencwajg Sep 24 '12 at 9:20