Why there is not nonconstant morphism from $P^{n}$ to $P^{1}$ when $ n>1$, and how about the case when we replace the $P^{1}$ by an arbitary projective curve $C$?
Another one: Let $C\subset P^{3}$ be a non-singular rational curve and $H$ be a hypersurface of degree two. How many points in the intersedtion set $C\bigcap H$?
There are no holomorphic maps from a projective space to a Kahler manifold X of smaller dimension. Indeed, the pullback of a Kahler form $\omega_X$ cannot be exact, because a positive closed form, integrated with an appropriate power of a Kahler form, is non-zero. Since the Picard group of P^n is Z, a pullback of a Kahler class is a Kahler class. However, $\omega_X^{dim X+1}=0$ downstairs, and upstairs it is non-zero because the cohomology algebra of P^n is truncated polynomials.
Such X is automatically Kahler, but it would take some effort to prove.
Geometric proof: For $\mathbf{P}^n \to \mathbf{P}^1$: Do you know the theorem on morphisms to projective space, [Hartshorne], Theorem II.7.1? Alternatively, one could use the "Projective Dimension Theorem", [Hartshorne], Theorem I.7.2.
For $\mathbf{P}^1 \to C$: Use the Hurwitz formula, [Hartshorne], Corollary IV.2.4.
For $\mathbf{P}^n \to C$: There is a $k = 2$ or $3$ such that there exists a closed immersion $C \hookrightarrow \mathbf{P}^k$. Similarly to the above, there are no non-constant morphisms $\mathbf{P}^n \to \mathbf{P}^k$ for $n > k$.
Cohomological proof: Alternatively: If $f: X \to Y$ is surjective, $f^\ast: H^\ast(Y) \hookrightarrow H^\ast(X)$ is injective for a Weil cohomology theory $H^\ast(-)$. For $\ell$-adic cohomology, $H^*_\ell(\mathbf{P}^n)$ can be easily calculated: $H_\ell^{2i}(\mathbf{P}^n_{\bar{k}}) = \mathbf{Q}_\ell$ for $i = 0, 1, \ldots, n$, and the other cohomology groups are $0$; and $H^1_\ell(C_{\bar{k}}) = \mathbf{Q}_\ell^{2g}$, $g$ the genus of $C$.
For the last question: [Hartshorne], Theorem I.7.7.
