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I have come across the following proposition (4.2.4 in Springer Linear algebraic groups, 1st ed.).

Proposition. Let $\phi \colon X \to Y$ be a dominant morphism of irreducible affine varieties. Let $x \in X$ be such that its fibre $\phi^{-1} (\phi (x))$ is finite. There is an affine open neighborhood $U$ of $\phi (x)$ in $Y$ such that $\phi^{-1} (U)$ is an affine open neighborhood of $x$ and that the restriction morphism $\phi^{-1} (U) \to U$ is finite.

I don't understand the proof, so I tried to see where the hypotheses are used in the following simple example.

Consider $\mathbb A^1 \times \mathbb A^1 \to \mathbb A^1$ sending $(x, y) \mapsto xy$. Now let $\Gamma = \{ (x, y, xy) \}$ be its graph and project the graph to the second two factors, leaving $\Gamma' = \{ (y, xy) \}$. Everything mentioned so far has been irreducible. Now consider the map $\phi \colon \Gamma' \to \mathbb A^1$ projecting to the first factor in the above lemma. The fibre over $y = 0$ is $(0, 0)$. The fibre over $y \neq 0$ is $\{ (y, xy) \vert x \in \mathbb A^1 \} \simeq \mathbb A^1$. However, an affine neighbourhood of the point $(0, 0)$ lifts to $\Gamma'$, but doesn't ever seem to be finite.

Still, this map seems to satisfy all the hypotheses of the above proposition. Is this a counterexample, or have I made a mistake somewhere?

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    $\Gamma^{\prime}$ is not a variety. – 2017-01-30
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    @ReneSchipperus I see. What condition do I need for $\Gamma'$ to be a variety? (Replacing $(x,y)\mapsto xy$ by a different map.) – 2017-01-30
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    Shafarevich I ch.I, 5.2 Thm 3 – 2017-01-30
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    @ReneSchipperus Thank you very much. – 2017-01-30
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    The result is wrong as it stands, unless there is a different definition of finite. Consider $X=\mathbb{C}-\{1}$, $Y=\mathbb{C}$, $\phi(x)=x^2$. Then $\phi(-1)=1$, but no such $U$ exists, which is an open neighbourhood of $1$. – 2017-01-31
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    @Mohan The definition given is "A dominant morphism $\phi \colon X \to Y$ of affine irreducible varieties is *finite* if $k [X]$ is an $k [Y]$-module of finite type." – 2017-01-31
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    @Mohan Taking $U = X$ and $\phi^{-1} (U) = Y$, the map $\mathbb C \setminus \{ 1 \} \to \mathbb C$ is not finite since $k [X] = k [x{-}1, (x{-}1)^{-1}]$ (Laurent polynomials) is not finite over $k [y]$. Is this what you are saying? – 2017-01-31
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    You also can not replace $U$ (you write $X$, but you mean $Y$, I think) with any open neighbourhood of $1$ and get a finite map. – 2017-01-31
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    Yes, sorry. I accidentally switched the labels of $X$ and $Y$ in my previous comment. – 2017-02-01

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