TITLE: Why a subvariety of a variety of general type is of general type
QUESTION [9 upvotes]: It seems a well-known fact that subvarieties of a variety of general type containing a general point are also of general type. This fact is an essential property used to prove some extension theorems of pluricanonical forms on algebraic varieties of general type, see for example the very nice survey on extension of pluricanonical forms
http://www.iecl.univ-lorraine.fr/~Gianluca.Pacienza/notes-grenoble.pdf (Internet Archive)
I want to know why this is true? Is there any thing more we can say about the relation between the canonical line bundle of a variety and that of subvarieties of codimension no less than 2.

REPLY [4 votes]: Theorem
Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$, then for general $t\in T$, let $X_{t}=f^{-1}(t)$, we have 
$$\kappa(X_{t}) \geq \kappa(X)-\dim T$$
proof:
For $m\in\mathbb{N}$ sufficiently large, consider Iitaka fibration $\phi=\phi_{mK_{X}}:X\dashrightarrow Z$, where $\dim Z=\kappa(X)$. Resolving the indeterminant locus of $\phi$, we may assume $\phi$ is a morphism.
\begin{align*}
    \kappa(X_{t})& \geq   \dim(\phi(X_{t}))\\
    &\dim X_{t}-\dim(X/Z)\\
    &=\dim Z-\dim T\\
    &=\kappa(X)-\dim T
\end{align*}
Corollary
Let $f:X\rightarrow T $ be a morphism of smooth projective varieties over $\mathbb{C}$. If $X$ is of general type, then any general fiber $X_{t}$ is of general type.
proof:\begin{align*}
    \kappa(X_{t})&\geq \kappa(X)-\dim T \\
    &=\dim X-\dim T\\ 
    &=\dim X_{t}\\
\end{align*}
The following Corollary tells us if we have a family of variety which dominants a variety of general type, then the general member of this family is also of general type.
Corollary Let $f: Z\rightarrow T$ be a projective morphism and $g:Z\rightarrow X$ is a dominant morphism to a variety of general type, then $Z_{t}$ is of general type for general $t\in T$.
proof:
By taking resolution of $X,Z,T$, we may assume that they are all smooth. Cutting by hyperplanes on $T$, we may assume that $\dim Z=\dim X$, hence $g$ is generically finite. We have 
$$K_{Z}=g^{*}K_{X}+R$$
where $R$ is effective. Since $K_{X}$ is big, it follows that $K_{Z}$ is big and $Z$ is of general type. So $Z_{t}$ is of general type for general $t\in T$.
Theorem Let $X$ be a projective variety of general type. $x\in X$ is a very general point. If $V$ is a subvariety containing $x$, then $V$ is also of general type.
proof:
The Hilbert scheme of $X$ $Hilb(X)=\bigcup_{p\in \mathbb{Q}[t]}Hilb_{P}(X)$ contained countably many components. For each $P\in \mathbb{Q}[t]$, we have the universal family $$U_{P}=Uni_{P}(X)\subset X\times Hilb_{P}(X)$$ Let $p:U_{P}\rightarrow X$ be the first projection. Then either the closure of $U_{P}$ is equal to $X$ or it's a proper closed set of $X$. Since our variety is over $\mathbb{C}$, an uncountable set, let $W$ be the union of all $\overline{U_{P}}$ such that $\overline{U_{P}}\neq X$, then $W\neq X$. For any $x\in X-W$, if $V$ is a subvariety containing $x$, then there is a universal family $U_{P}$ for some $P$, such $V$ is a fiber of $U_{P}\rightarrow Hilb_{P}(X)$ and $p:U_{P}\rightarrow X$ is dominant. Then the assertion follows form the Corollary above.