Consider the code:

#include <iostream>
#include <vector>
#include <ranges>

class TA
{
private:
    int  value;
public:
    TA(int _value) : value(_value)
    {
        std::cout << "Constructor TA, value = " << value << 'n';
    }
    TA(const TA& a)
    {
        value = a.value;
        std::cout << "Copy constructor TA, value = " << value << 'n';
    }
    TA(TA&& a) noexcept
    {
        value = a.value;
        std::cout << "Move constructor TA, value = " << value << 'n';
    }
    ~TA()
    {
        std::cout << "Destructor TA, value = " << value << 'n';
    }
};

void rvalue_vector_test1()
{
    std::vector a{ std::from_range, std::vector{TA(1), TA(2), TA(3), TA(4), TA(5)} };
}

void rvalue_vector_test2()
{
    std::vector a{ std::from_range, std::vector{TA(1), TA(2), TA(3), TA(4), TA(5)} | std::views::as_rvalue };
}

void iota_test1()
{
    std::vector a{ std::from_range,
                   std::views::iota(1,6) |
                   std::views::transform([](auto elem) {return TA(elem); }) };
}


int main()
{
   rvalue_vector_test1();
   rvalue_vector_test2();
   iota_test1();
}

Could anybody explain why in the code:
std::vector a{ std::from_range, std::vector{TA(1), TA(2), TA(3), TA(4), TA(5)} | std::views::as_rvalue }
it is necessary to use std::views::as_rvalue to prevent copying (although std::vector{TA(1), TA(2), TA(3), TA(4), TA(5)} is rvalue), but in the code:

std::vector a{ std::from_range,
               std::views::iota(1,6) |
               std::views::transform([](auto elem) {return TA(elem); }) };

there is no need to use std::views::as_rvalue

(P.S. I know that the best way is to write:

std::vector<TA> a{ std::from_range, std::views::iota(1,6) }

– there is no copying and moving at all, but I want to know, how constructors with std::from_range work)