can std::forward takes std::function

I've always seen std::forward being utilized as below, utilized inside a template function

template<class T>
void foo(T&& arg): bar(std::forward<T>(arg)){}

Suppose I want to do this.

class A
{
private:
  std::function<void()> bar;
public:
  template<class T>
  void foo(T&& arg): bar(std::forward<T>(arg)){}
}

Since bar already has its type defined. Can I rewrite it by directly specifying T as std::function >? Compiler doesn't seem happy about it. Why?

class A
{
private:
  std::function<void()> bar;
public:
  void foo(std::function<void()>&& arg): bar(std::forward<std::function<void()>>(arg)){}
    }