Is the restriction of a diffeomorphism to a manifold $M$ an open map from $M$ to $M$? Let $d\in\mathbb N$, $k\in\{1,\ldots,d\}$, $M$ be a $k$-dimenional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary, $T$ be a $C^1$-diffeomorphism of $\mathbb R^d$ onto $\mathbb R^d$ and $N:=T(M)$.
We know that the restriction of a homeomorphism is a homeomorphism and hence $\left.T\right|_M$ is a homeomorphism of $M$ onto $N$. In particular, $T$ is an open map from $M$ to $N$.

Question 1: Assuming $T(M)\subseteq M$, are we even able to show that $\left.T\right|_M$ is an open map from $M$ to $M$?
Question 2: In the same spirit, are we able to show $\left.T\right|_{\partial M}$ is an open map from $\partial M$ to $\partial M$?

 A: Since the symbol $T$ is usually used for the pushforward and tangent space, let me replace it by $f$.
We need to assume that $\partial M=\emptyset$.
Now let $\Omega$ be an open subset of $M$. We need to show that $\Omega':=f(\Omega)$ is $M$-open. Let $x\in\Omega$. By assumption, $\Omega'\subseteq f(M)\subseteq M$ and hence $$T_{f(x)}\:\Omega'\subseteq T_{f(x)}\:M\tag1.$$ Since $f$ is a $C^1$-diffeomorphism, $\Omega'$ is a $k$-dimensional $C^1$-submanifold of $\mathbb R^d$ as well and hence $$\dim T_{f(x)}\:\Omega'=k=\dim T_{f(x)}\:M\tag2.$$ So, $(1)$ is actually an equality. Since $\left.f\right|_\Omega$ is a $C^1$-diffeomorphism from $\Omega$ onto $\Omega'$, this implies that $T_x(\left.f\right|_\Omega)$ is an isomorphism from $T_x\:\Omega$ onto $T_{f(x)}\:\Omega'=T_{f(x)}\:M$. From a version of the inverse function theorem, we can infer that $\left.f\right|_{\tilde\Omega}$ is a $C^1$-diffeomorphism from $\tilde\Omega$ onto $\tilde\Omega':=f(\tilde\Omega)$ for some $\Omega$-open neighborhood $\tilde\Omega$ of $x$ and $\tilde\Omega'$ is an $M$-open neighborhood of $f(x)$. By construction, $$\tilde\Omega'\subseteq\Omega'\tag3.$$ Since $x$ was arbitary, $\Omega'$ is $M$-open and hence we have shown that $\left.f\right|_M$ is an open map from $M$ to $M$.
