This question was previously posted on MSE.

Let $M, N$ be smooth connected manifolds (without boundary), where $M$ is a compact manifold, so we can put a topology in the space $\mathcal C^\infty(M, N)$ using $\mathcal{C}^1$ Whitney Topology.

Now, consider $S\subset M$ a compact submanifold of $M$ with boundary such that $\text{dim}S=\text{dim}M$, using the same process we can put a topology in $\mathcal C^\infty(S,N)$ using the $\mathcal{C}^1$ Whitney Topology. There is a natural continous projection of $\mathcal C^\infty(M, N)$ on $\mathcal C^\infty(S,N)$, definided by

\begin{align*} \pi: \mathcal C^\infty(M, N) &\to \mathcal C^\infty(S,N)\\ f&\mapsto \left.f\right|_{S}. \end{align*}

My Question:Is $\pi$ an open map or at least a quotient map?

## Some comments

$\mathcal{C}^1$-Whitney Topology is also called $\mathcal{C}^1$-strong topology.

As noticed for the user Adam Chalumeau, on the book "Morris W. Hirsh Differential Topology" there is the following exercise

[Exercise 16, page 41]:Let $M, N$ be $\mathcal{C}^r$ manifolds. Let $V⊂M$ be an open set then

The restriction map $$δ:\mathcal{C}^r(M,N)→\mathcal{C}^r(V,N)$$ $$δ(f)=f|_V$$ is continuous for the weak topology, but not always for the strong.

$δ$ is open for the strong topologies, but not always for the weak".

Since our $M$ is compact weak topology = strong topology. However, I don't know how to solve this exercise let alone adapt such proof to the case that I want.

Does anyone know anything about this problem?