Downcast a std::unique_ptr to std::unique_ptr

This question already has an answer here:

• “Downcasting” unique_ptr<Base> to unique_ptr<Derived> 1 answer

What is the appropriate way to downcast a std::unique_ptr ? There's these other question that asks the same but I am unsure if it is under the OP's specific context or if that applies to the general case.

In that question, this is the solution:

template<typename Derived, typename Base, typename Del>
std::unique_ptr<Derived, Del> 
static_unique_ptr_cast( std::unique_ptr<Base, Del>&& p )
{
    auto d = static_cast<Derived *>(p.release());
    return std::unique_ptr<Derived, Del>(d, std::move(p.get_deleter()));
}

But I've also seen only:

template<typename D, typename B>
std::unique_ptr<D> static_unique_ptr_cast(std::unique_ptr<B>& base)
{
    return std::unique_ptr<D>(static_cast<D*>(base.release()));
}

Is there a good way to do this?

Do we need to move the deleter in general?

What happens if we do:

std::unique_ptr ptr = MakeDerived();
static_unique_ptr_cast(ptr)->DoSomething();

And why doesn't the standard provide this capability?