TITLE: Do locally convex topological vector spaces embed into diffeological spaces?
QUESTION [11 upvotes]: The nLab casually remarks that locally convex tvs embed into diffeological spaces by (discussion around) a corollary in Kriegl and Michor, namely 3.14, but this deals with Boman's theorem and results about smooth maps from a euclidean space to a locally convex spaces.
The definition of smoothness for maps $U \to V$ where $U$ is $c^\infty$-open in a lctvs $W$, and $V$ is a lctvs, is given so that smooth maps are automatically the same as between maps between the associated diffeological spaces.
However, there is a notion of smoothness for maps between lctvs (due to Michal [1,2] and Bastiani [3]) not relying on K&M's definition, but I don't how they relate (Keller [4] treats this, I think, but I don't have access to that book at the moment).

Is it still true that lctvs embed into diffeological spaces using Michal-Bastiani smoothness?

[1] Michal, A. D. Differential calculus in linear topological spaces, Proc. Nat. Acad. Sci. USA 24 (1938), 340-342 (pdf)
[2] Michal, A. D. Differential of  functions with arguments and values in  topological  abelian groups, Proc. Nat. Acad. Sci. USA 26 (1940), 356–359. (pdf)
[3] Bastiani, A., Applications différentiables et variétés différentiables de dimension infinie, J. Anal. Math. 13 (1964), 1–114. (article, paywalled)
[4] Keller, H. H., Differential Calculus in Locally Convex Spaces, Lecture Notes in Mathematics 417, Springer-Verlag, 1974 (Springerlink, paywalled)

REPLY [3 votes]: $\def\sp{\kern.4mm}\def\bbN{\mathbb N}\def\bbZ{\mathbb Z}\def\bbR{\mathbb R}$Here is the answer to David Roberts' residual query: is a convenient diffeomorphism necessarily MB-smooth? No. A counterexample follows:
Let $E$ be the vector space of all real (two-sided) sequences $x=\langle\sp x_i:i\in\bbZ\sp\rangle$ for which $x_i=0$ for sufficiently small $i$ , topologized 
so that we get a linear homeomorphism $E\to\bbR\,^{\bbN}\times\bbR\,^{(\bbN)}$ defined by $x\mapsto(u,v)$ where $u_i=x_{i-1}$ and $v_i=x_{-i}$ for $i\in\bbN$ . Then define $f:E\to E$ by $x\mapsto y$ where $y_i=x_i$ for $i\in\bbZ\setminus\{\sp 0\sp\}$ and $y_0=x_0+\sum_{i\in\bbN}(x_i\cdot x_{-i})$ . The inverse is given by $y\mapsto x$ where $y_i=x_i$ for $i\in\bbZ\setminus\{\sp 0\sp\}$ and $x_0=y_0-\sum_{i\in\bbN}(y_i\cdot y_{-i})$ . Since the duality map $\bbR\,^{\bbN}\times\bbR\,^{(\bbN)}\to\bbR$ is a discontinuous but bornological, and hence conveniently smooth bilinear map, the assertion follows.