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.



Pull-back of generalized functions
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.


