Pull-back of generalized functions

Question

Let $f\colon X\to Y$ be a smooth map between smooth manifolds. Then the pull-back operation $f^*\colon C^\infty(Y)\to C^\infty(X)$ is a linear continuous operator when $C^\infty$ is equipped with the usual topology of of uniform convergence on compact subsets of all partial derivatives.

It tuns out that sometimes $f^*$ can be extended to larger spaces of generalized functions. More precisely, fix a closed $\mathbb{R}_{>0}$-invariant subset $\Lambda\subset T^*Y\backslash 0$ of the cotangent bundle with removed zero section. Let $C^{-\infty}_\Lambda(Y)$ denote the space of generalized functions with the wave front set contained in $\Lambda$. This space $C^{-\infty}_\Lambda(Y)$ is equipped with some standard locally convex linear topology (see e.g. Ch. 6 in the book "Geometric Asymptotics" by Guillemin and Sternberg).

Let us assume that the map $f$ is transversal to $\Lambda$ in the sense that for any $x\in X$ if $(f(x),\eta)\in \Lambda$ then $(df_x^*)(\eta)\ne 0$. Then one defines a ("natural") linear map

$$f^*\colon C^{-\infty}_\Lambda(Y)\to C^{-\infty}(X)\,\,\,\,\,\,\,\,\,\, (1)$$

extending the usual pull-back on smooth functions (see the above mentioned book. Another very good reference is Hormander's ”The analysis of linear partial differential operators, I”; see especially Theorem 8.2.4.)

The point is that in the above literature the map (1) is proven to be sequentially continuous, namely it maps convergent sequences to convergent ones.

QUESTION. Is the pull-back map (1) topologically continuous?

The difference between usual topological continuity (e.g. continuity in the usual sense of maps of topological spaces) and sequential continuity seems to me to be quite subtle.

Edit: Definition of topology on $C^{-\infty}_\Lambda(X)$. Covering $X$ by open charts and using partition of unity we may assume that $X= \mathbb{R}^n$. For any $N\in\mathbb{N}, \phi \in C^\infty_c(\mathbb{R}^n)$, and any closed $\mathbb{R}_{>0}$-invariant subset $V\subset \mathbb{R}^n$ such that $$\Lambda\cap (supp(\phi)\times V)=\emptyset$$ let us define the semi-norm on $C^{-\infty}_\Lambda(\mathbb{R}^n)$ by $$||u||_{\phi,N,V}=sup_{\xi\in V}|\xi|^N|\widehat{\phi u}(\xi)|,$$ where $\hat F$ denotes the Fourier transform of the function $F$. Then one equips $C^{-\infty}_\Lambda(X)$ with the weakest locally convex topology which is stronger than the weak topology on $C^{-\infty}(X)$ and such that all semi-norms $\{||\cdot||_{\phi,N,V}\}$ are continuous.

Answer 1

Recently I found a counterexample to a somewhat stronger version of the question I originally asked, but which I actually needed. Namely for a smooth map $f\colon X\to Y$ which is transversal to $\Lambda\subset T^*Y\backslash 0$ (see e.g. Hormander's book for the definitions) the map $f^*\colon C^{-\infty}_{\Lambda}(Y)\to C^{-\infty}_{f^*\Lambda}(X)$ may not be topologically continuous (notice the subscript $f^*\Lambda$ in the target).

The counterexample is quite simple. Let $f\colon \mathbb{R}^2\to \mathbb{R}$ be the projection to the first coordinate. Let $\Lambda=\mathbb{R}\times (\mathbb{R}\backslash 0)$. Then $C^{-\infty}_{\Lambda}(\mathbb{R})=C^{-\infty}(\mathbb{R})$ is the usual Schwartz space with the usual weak topology. Furthermore $f^*\Lambda=\{(x,y;\xi,0)\}$. Then the map $f^*\colon C^{-\infty}_\Lambda(\mathbb{R})\to C^{-\infty}_{f^*\Lambda}(\mathbb{R}^2)$ is not topologically continuous.

Strictly speaking I do not yet have a counterexample to my original question, but this one already suffices my purposes.

Answer 2

I deleted my previous answer since "googling" the book "Geometric Asymptotics" by Guillemin and Sternberg revealed that my provisions there were not in accordance with the presentation in the book. Chapter VI begins on page 305, and from page 306 it becomes plausible (but not clear) that by "$C^{-\infty}$" the authors refer to the dual of $C_0^{\,\infty}$ and not to that of $C^{\,\infty}$ which I assumed earlier. Unfortunately, I couldn't see page 333, but Proposition 3.7 on page 335 claims that $f^*v$ for $f:X\to Y$ can be resonably defined as a certain limit of a sequence when $v$ is a distribution in $C_Z^{-\infty}(Y)$ . Here $Z$ is the $\Lambda$ of OP. I didn't find any claim of this extended pull-back operation being even sequentially continuous $C_Z^{-\infty}(Y)\to C^{-\infty}(X)$ .

However, having thought more of the problem, it has began to seem to me that continuity might be able to be proven in the case where $X$ and $Y$ have the same dimension and $f$ is "almost" a diffeomorphism. When the dimensions are not equal, it seems that one only can prove that for $f^*:C^{\,\infty}(Y)\to C^{\,\infty}(X)$ its restriction to every bounded set in $C^{-\infty}(Y)$ is continuous when the spaces $C^{\,\infty}$ are equipped with the initial topologies by the embeddings $C^{\,\infty}(Y)\to C_Z^{-\infty}(Y)$ and $C^{\,\infty}(X)\to C^{-\infty}(X)$ . This allows the extended pull-back to be reasonably defined, and possibly also sequential continuity (but not continuity) of $f^*:C_Z^{-\infty}(Y)\to C^{-\infty}(X)$ to be established.