Why is template argument deduction disabled with std::forward?

In VS2010 std::forward is defined as such:

template<class _Ty> inline
_Ty&& forward(typename identity<_Ty>::type& _Arg)
{   // forward _Arg, given explicitly specified type parameter
    return ((_Ty&&)_Arg);
}

identity appears to be used solely to disable template argument deduction. What's the point of purposefully disabling it in this case?