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 quocient 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?