Proposed resolution:

Change [forward] as indicated:

template <class T> T&& forward(typename identity<T>::type&& t);

...

Return type: If T is an lvalue-reference type, an lvalue; otherwise, an rvalue.

...

-7- In the first call to factory, A1 is deduced as int, so 2 is forwarded to A's constructor as an int&& (an rvalue). In the second call to factory, A1 is deduced as int&, so i is forwarded to A's constructor as an int& (an lvalue). In both cases, A2 is deduced as double, so 1.414 is forwarded to A's constructor as double&& (an rvalue).

template <class T> typename remove_reference<T>::type&& move(T&& t);

...

Return type: an rvalue.

p4 (forward) says:

Return type: If T is an lvalue-reference type, an lvalue; otherwise, an rvalue.

First of all, lvalue-ness and rvalue-ness are properties of an expression, not of a type (see [basic.lval]). Thus, the phrasing "Return type" is wrong. Second, the phrase says exactly what the core language wording says for folding references in [temp.arg.type]/p4 and for function return values in [expr.call]/p10. (If we feel the wording should be retained, it should at most be a note with cross-references to those sections.)

The prose after the example talks about "forwarding as an int& (an lvalue)" etc. In my opinion, this is a category error: "int&" is a type, "lvalue" is a property of an expression, orthogonal to its type. (Btw, expressions cannot have reference type, ever.)

Similar with move:

Return type: an rvalue.

is just wrong and also redundant.