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Is the following assertion true?

"Let $X$ and $Y$ be two quasi-projective varieties and $f:X \longrightarrow Y$ a morphism which is injective and dominant. Then, the dimensions of $X$ and $Y$ coincide."

Certainly, if $f$ is just dominant, we have that $$\dim X \geq \dim Y,$$ as was shown in the answer to this question. So the problem is if equality holds if we impose injectivity.

If I am not wrong, we could deduce the result if $f$ were closed, by using the topolgical definition of dimension as in Hartshorne's book. However, since $X$ is not necessarily a projective variety but a quasi-projective, I guess we cannot use this argument.

Any help or suggestion is welcome.

Don
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  • Given a dominant morphism $f:X\to Y$ of quasi-projective varieties, there exists a non-empty open subset $U$ of $Y$ such that $U\subset f(X)$. So, in particular, the map $f^{-1}(U)\to U$ is onto. So, with your hypothesis on injectivity, you should be able to finish the argument. – Mohan Jun 20 '18 at 14:42
  • @Mohan Clearly, the map $f^{-1}(U) \longrightarrow U$ is a bijective morphism between quasi-projective varieties. However, if I am not wrong, we cannot deduce from here that the map admits a morphism as inverse; that is, we cannot deduce that it is an isomorphism. In fact, I even doubt it is birational, which is actually the case in which we are interested. Am I wrong? – Don Jun 20 '18 at 16:44
  • No, you can not deduce it is an isomorphism, but, you can assume further, that the map is finite (usually called generic finiteness) and thus the dimensions are the same. – Mohan Jun 20 '18 at 17:06
  • Dear @Mohan, I am very grateful for your answers. Many thanks. – Don Jun 20 '18 at 23:01

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