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An Interesting Exercise

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An Interesting Exercise
Let \( \displaystyle X \) be a compact subspace of \( \displaystyle \mathbb{R}^{n} \). Consider the rings \[ \displaystyle C \left( X , \mathbb{R} \right) \, , \, C \left( \mathbb{R}^{n} , \mathbb{R} \right) \] of continuous functions and define the mapping \[ \displaystyle \Phi : C \left( \mathbb{R}^{n} , \mathbb{R} \right) \longrightarrow C \left( X , \mathbb{R} \right) \, , \, \Phi(f) = f_{X} \] Show that \[ \displaystyle C \left( \mathbb{R}^{n} , \mathbb{R} \right) / Ker(\Phi) \cong C \left( X , \mathbb{R} \right) \]

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Re: An Interesting Exercise
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.
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.

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Re: An Interesting Exercise
1) If X is compact then is not subspace
2)The functions in HahnBanach theorem are linear
2)The functions in HahnBanach theorem are linear

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Re: An Interesting Exercise
1) As $X$ is a subset of $ \mathbb{R}^{n} $, it is naturally a subspace; We additionally require it to be compact.S.F.Papadopoulos wrote: 1) If X is compact then is not subspace
2)The functions in HahnBanach theorem are linear
2)Hahn Banach Theorem/ Formulation
What can you say about the solution of the exercise? How is your post related to it? Could you write a few more things and be more specific?

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Re: An Interesting Exercise
1) subspace of a vector space for me( Rudin Functional Analysis and other) mean linear subspace.
2)This is a simple version of Tietze Extension Theorem.
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2)This is a simple version of Tietze Extension Theorem.
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