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I don't think that any algebraic variety can be isomorphic to a proper open subset. Over a finite field this follows immediately from counting points over extensions. In the general case one reduces t …

Yes, both results are true. For the first, as you say, the image of the map is either a point, or dominates an affine curve $C$ of geometric genus 0. By standard results, it factors through the normal …

The only smooth affine curve admitting a non-constant map from an affine space is $\mathbb A^1$. It must have genus 0, because of Lüroth's theorem, so it is $\mathbb P^1$ minus $d$ points for some $d$ …

It is easy to see that $D + \tau^*D$ is a canonical divisor. Suppose that $p$ is a ramification point of the $g^{1}_{2}$; since there are $2g-2$ such points, we may assume that $p$ is not a base point …

The points of $X^{(2)}$ correspond to divisors of degree 2; two points of $X^{(2)}$ are in the image of a morphism $\mathbb P^1 \to X^{(2)}$ if and only if the two corresponding divisors are linearly …

I don't think this is true. Take $X = \mathbb P^1 \times \mathbb P^2$. Let $C$ be $\mathbb P^1 \times 0$, and let $C_1$ be the first infinitesimal neighborhood of $C$. The curve $C_1$ is the relative …

The splitting theorem is most certainly false for vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$. In fact, the theory of vector bundles on quadric surfaces is probably as complicated as the theory …