Why do not use universal reference to implement std::forward?

I knew there were probably already some similar questions on StackOverflow, but I didn't actually find anything that fully answered my doubts, so I asked my own question.

There are two different overloads of std::forward:

template< class T >
T&& forward( typename std::remove_reference<T>::type& t );

template< class T >
T&& forward( typename std::remove_reference<T>::type&& t );

What is the difference with one universal reference template function using static_cast? Like:

template<class T>
T&& myForward(T &&t) {
    return static_cast<T&&>(t);
}

When I pass a rvalue of type A, T with be deduced to A and myForward will return A &&; When a lvalue is passed, T with be deduced to A &, and it will return A &. So what is the difference?

I understand the notion that the second overload can prohibit forwarding a rvalue to a lvalue, but why myForward cannot?