Firstly, the mapping \(\displaystyle{\Phi}\) is well defined because, if \(\displaystyle{f\in C(\mathbb{R}^n,\mathbb{R})}\),
then \(\displaystyle{f_{X}}\) is continuous, that is \(\displaystyle{f_{X}\in C(X,\mathbb{R})}\).
If \(\displaystyle{f\,,g\in C(\mathbb{R}^n,\mathbb{R})}\) and \(\displaystyle{a\in\mathbb{R}}\), then,
\(\displaystyle{\Phi(f+a\,g)=(f+a\,g)_{X}=f_{X}+a\,g_{X}=\Phi(f)+a\,\Phi(g)}\)
\(\displaystyle{\Phi(f\,g)=(f\,g)_{X}=f_{X}\,g_{X}=\Phi(f)\,\Phi(g)}\)
and \(\displaystyle{\Phi(\mathbb{1})=\mathbb{1}_{X}=\mathbb{1}}\).
So, the mapping \(\displaystyle{\Phi}\) is an \(\displaystyle{\mathbb{R}}\) algebra homomorphism.
Let \(\displaystyle{g\in C(X,\mathbb{R})}\). Since, \(\displaystyle{X}\) is subspace of \(\displaystyle{\mathbb{R}^n}\),
and according to \(\displaystyle{\rm{HahnBanach}}\)  theorem, there exists \(\displaystyle{f\in C(\mathbb{R}^n,\mathbb{R})}\)
such that \(\displaystyle{f_{X}=g\iff \Phi(f)=g}\). Therefore, the mapping \(\displaystyle{\Phi}\)
is onto.
According to the 1st Isomorphism Theorem we have that
\(\displaystyle{C(\mathbb{R}^n,\mathbb{R})/\rm{Ker}(\Phi)\cong C(X,\mathbb{R})}\) as \(\displaystyle{\mathbb{R}}\)  algebras.
