# If $\phi:V_1\rightarrow V_2$ is a morphism of varieties then $V_1\cong \phi(V_1)$

I am reading Silverman's The Arithmetic of Elliptic Curves. I am wondering if with the definition of morphism he gives, we can conclude that if $$\phi:V_1\rightarrow V_2$$ is a morphism of projective varieties then $$V_1\cong \phi(V_1)$$. Throughout my mathematical career, I have seen cases where a morphism induces an isomorphism to the image, and cases where it doesn't. It usually depends on how restrictive you want the morphism condition to be. I don't really know what happens in this case, and would appreciate some insight. Here are the definitions Silverman uses for (iso)morphism of projective varieties:

Let $$V_1$$ and $$V_2 \subset \mathbb{P}^n$$ be projective varieties. A rational map from $$V_1$$ to $$V_2$$ is a map of the form $$\phi: V_1 \longrightarrow V_2, \quad \phi=\left[f_0, \ldots, f_n\right],$$ where the functions $$f_0, \ldots, f_n \in \bar{K}\left(V_1\right)$$ have the property that for every point $$P \in V_1$$ at which $$f_0, \ldots, f_n$$ are all defined, $$\phi(P)=\left[f_0(P), \ldots, f_n(P)\right] \in V_2 .$$

A rational map $$\phi=\left[f_0, \ldots, f_n\right]: V_1 \longrightarrow V_2$$ is regular (or defined) at $$P \in V_1$$ if there is a function $$g \in \bar{K}\left(V_1\right)$$ such that (i) each $$g f_i$$ is regular at $$P$$; (ii) there is some $$i$$ for which $$\left(g f_i\right)(P) \neq 0$$.

A rational map that is regular everywhere is called a morphism. Finally:

Let $$V_1$$ and $$V_2$$ be varieties. We say that $$V_1$$ and $$V_2$$ are isomorphic, and write $$V_1 \cong V_2$$, if there are morphisms $$\phi: V_1 \rightarrow V_2$$ and $$\psi: V_2 \rightarrow V_1$$ such that $$\psi \circ \phi$$ and $$\phi \circ \psi$$ are the identity maps on $$V_1$$ and $$V_2$$, respectively.

• No, consider $P^n\times P^m \to P^m, (x, y) \to y$ for example, with $P^n\times P^m$ projective as it can be embedded into some $P^N$. Commented Mar 9 at 13:50

Let $$E\subset\Bbb P^2$$ be an elliptic curve in Weierstrass form. Then the projection to the $$x$$-axis gives a 2-to-1 surjective map from $$E\to\Bbb P^1$$, and $$E\not\cong\Bbb P^1$$ since they have different genera.
Even if you require bijectivity, this won't hold: consider the normalization map $$\Bbb P^1\to V(y^2z-x^3)\subset\Bbb P^2$$, which is bijective but not an isomorphism, since the source is smooth but the target is singular.