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](http://web.archive.org/web/20190320105238/http://www.iecl.univ-lorraine.fr/~Gianluca.Pacienza/notes-grenoble.pdf))


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. 



Why a subvariety of a variety of general type is of general type
Let me reproduce the relevant bit from the [reference](http://www.iecl.univ-lorraine.fr/~Gianluca.Pacienza/notes-grenoble.pdf) above ([Internet Archive](http://web.archive.org/web/20190320105238/http://www.iecl.univ-lorraine.fr/~Gianluca.Pacienza/notes-grenoble.pdf)):



> 
> **Exercise 3.1.** Let $X$ be a variety of general type. Prove that a
>  subvariety of $X$ passing through a
>  general point is of general type.
> 
> 
> 


Bogomolov proved in his paper "Families of curves on a surface of general type" that surfaces of general type satisfying $c\_1^2 > c\_2$ contain at most a finite number of rational and elliptic curves. It is unknown if the same holds true for an arbitrary surface of general type. Thus in the exercise above general point does not mean outside a closed subvariety. Instead it means outside a countable union of closed subvarieties. Of course, the author is considering all his varieties defined over $\mathbb C$.


Standard arguments reduces the exercise to the following



> 
> **Exercise.** Let $X$ be a variety of
>  general type. If $Y \rightarrow T$ is a family
>  of irreducible general type subvarieties of $X$ parametrized by
>  a irreducible complex variety $T$ then the
>  natural map $Y\to X$ is not dominant.
> 
> 
> 


