Proof.
We will use the results of Morphisms, Sections 29.18, 29.52, and 29.54 without further mention. We may replace Y by its normalization. Let X_0 \to X be the normalization. The morphism Y \to X factors through X_0. Thus we may assume that both X and Y are normal.
Assume X and Y are normal. The morphism f : Y \to X is an isomorphism over an open which contains every point of codimension 0 and 1 in Y and every point of Y over which the fibre is finite, see Varieties, Lemma 33.17.3. Hence there is a finite set of closed points T \subset X such that f is an isomorphism over X \setminus T. For each x \in T the fibre Y_ x is a proper geometrically connected scheme of dimension 1 over \kappa (x), see More on Morphisms, Lemma 37.53.6. Thus
BadCurves(f) = \{ C \subset Y\text{ closed} \mid \dim (C) = 1, f(C) = \text{a point}\}
is a finite set. We will prove the lemma by induction on the number of elements of BadCurves(f). The base case is the case where BadCurves(f) is empty, and in that case f is an isomorphism.
Fix x \in T. Let X' \to X be the normalized blowup of X at x and let Y' be the normalization of Y \times _ X X'. Picture
\xymatrix{ Y' \ar[r]_{f'} \ar[d] & X' \ar[d] \\ Y \ar[r]^ f & X }
Let x' \in X' be a closed point lying over x such that the fibre Y'_{x'} has dimension \geq 1. Let C' \subset Y' be an irreducible component of Y'_{x'}, i.e., C' \in BadCurves(f'). Since Y' \to Y \times _ X X' is finite we see that C' must map to an irreducible component C \subset Y_ x. If is clear that C \in BadCurves(f). Since Y' \to Y is birational and hence an isomorphism over points of codimension 1 in Y, we see that we obtain an injective map
BadCurves(f') \longrightarrow BadCurves(f)
Thus it suffices to show that after a finite number of these normalized blowups we get rid at of at least one of the bad curves, i.e., the displayed map is not surjective.
We will get rid of a bad curve using an argument due to Zariski. Pick C \in BadCurves(f) lying over our x. Denote \mathcal{O}_{Y, C} the local ring of Y at the generic point of C. Choose an element u \in \mathcal{O}_{X, C} whose image in the residue field R(C) is transcendental over \kappa (x) (we can do this because R(C) has transcendence degree 1 over \kappa (x) by Varieties, Lemma 33.20.3). We can write u = a/b with a, b \in \mathcal{O}_{X, x} as \mathcal{O}_{Y, C} and \mathcal{O}_{X, x} have the same fraction fields. By our choice of u it must be the case that a, b \in \mathfrak m_ x. Hence
N_{u, a, b} = \min \{ \text{ord}_{\mathcal{O}_{Y, C}}(a), \text{ord}_{\mathcal{O}_{Y, C}}(b)\} > 0
Thus we can do descending induction on this integer. Let X' \to X be the normalized blowing up of x and let Y' be the normalization of X' \times _ X Y as above. We will show that if C is the image of some bad curve C' \subset Y' lying over x' \in X', then there exists a choice of a', b' \mathcal{O}_{X', x'} such that N_{u, a', b'} < N_{u, a, b}. This will finish the proof. Namely, since X' \to X factors through the blowing up, we see that there exists a nonzero element d \in \mathfrak m_{x'} such that a = a' d and b = b' d (namely, take d to be the local equation for the exceptional divisor of the blowup). Since Y' \to Y is an isomorphism over an open containing the generic point of C (seen above) we see that \mathcal{O}_{Y', C'} = \mathcal{O}_{Y, C}. Hence
\text{ord}_{\mathcal{O}_{Y, C}}(a) = \text{ord}_{\mathcal{O}_{Y', C'}}(a' d) = \text{ord}_{\mathcal{O}_{Y', C'}}(a') + \text{ord}_{\mathcal{O}_{Y', C'}}(d) > \text{ord}_{\mathcal{O}_{Y', C'}}(a')
Similarly for b and the proof is complete.
\square
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