Let $\Omega\subset \mathbb{R}^n$ be a bounded open set with $C^1$ boundary. Consider the space $$\mathbb{X}(\Omega):=\left\{u \in H^1\left(\mathbb{R}^n\right): u \equiv 0 \text { in } \mathbb{R}^n \backslash \Omega\right\} .$$ The paper I am reading says that
We observe that, in view of the regularity assumption on $\Omega$, the space $\mathbb{X}(\Omega)$ is contained in $H^1\left(\mathbb{R}^n\right)$ and is isomorphic to $H^1_0\left(\Omega\right)$ via the ‘zero-extension’ map defined as $$\mathcal{E}_0: H_0^1\left(\mathbb{R}^n\right) \rightarrow \mathbb{X}(\Omega), \quad \text { such that } \quad \mathcal{E}_0(u):=u \cdot \chi_{\Omega}.$$
It is easy to see that if $u\in H_0^1(\Omega)$ then its zero-extension is an element of $\mathbb{X}(\Omega)$.
I tried to show that for all $v\in\mathbb{X}(\Omega)$, there exists $u\in H^1_0(\Omega)$ whose zero-extension is $v$. That is, $v|_\Omega\in H^1_0(\Omega), \forall v\in\mathbb{X}(\Omega)$. But I do not have any idea.