How does Gson TypeToken work?

From §4.6 of the JLS (emphasis mine):

Type erasure is a mapping from types (possibly including parameterized types and type variables) to types (that are never parameterized types or type variables). We write |T| for the erasure of type T. The erasure mapping is defined as follows:

The erasure of a parameterized type (§4.5) G is |G|.

The erasure of a nested type T.C is |T|.C.

The erasure of an array type T[] is |T|[].

The erasure of a type variable (§4.4) is the erasure of its leftmost bound.

The erasure of every other type is the type itself.

Therefore, if you declare a class with an anonymous subclass of itself, it keeps it's parameterized type; it's not erased. Therefore, consider the following code:

import java.lang.reflect.ParameterizedType;
import java.util.Arrays;
import java.util.HashMap;

public class Erasure<T>
{
    public static void main(String...strings) {
      Class<?> foo = new Erasure<HashMap<Integer, String>>() {}.getClass();
      ParameterizedType t = (ParameterizedType) foo.getGenericSuperclass();
      System.out.println(t.getOwnerType());
      System.out.println(t.getRawType());
      System.out.println(Arrays.toString(t.getActualTypeArguments()));
    }
}

This outputs:

null
class Erasure
[java.util.HashMap<java.lang.Integer, java.lang.String>]

Notice that you would get a if you did not declare the class anonymously, because of erasure; the superclass would not be a parameterized type, it would be an .ClassCastExceptionObject

Java's type erasure applies to individual objects, not classes or fields or methods. TypeToken uses an anonymous class to ensure it keeps generic type information, instead of just creating an object.