Meaning of std::forward

I have a question about the code written in https://koturn.hatenablog.com/entry/2018/06/10/060000 When I pass a left value reference, if I do not remove the reference with std::decay_t, I get an error.
Here is the error message
'error: 'operator()' is not a member of 'main()::<lambda(auto:11, int)>&
I don't understand why it is necessary to exclude the left value reference.
I would like to know what this error means.

#include <iostream>
#include <utility>


template <typename F>
class
FixPoint : private F
{
public:
  explicit constexpr FixPoint(F&& f) noexcept
    : F{std::forward<F>(f)}
  {}

  template <typename... Args>
  constexpr decltype(auto)
  operator()(Args&&... args) const
  {
    return F::operator()(*this, std::forward<Args>(args)...);
  }
};  // class FixPoint


namespace
{
template <typename F>
inline constexpr decltype(auto)
makeFixPoint(F&& f) noexcept
{
  return FixPoint<std::decay_t<F>>{std::forward<std::decay_t<F>>(f)};
}
}  // namespace


int
main()
{
  auto body = [](auto f, int n) -> int {
    return n < 2 ? n : (f(n - 1) + f(n - 2));
  };
  auto result = makeFixPoint(body)(10);
  std::cout << result << std::endl;
}