Let $C(\mathbb{T})$ denotes the C* algebra of continuous functions on the unit circle. Let $\alpha: C(\mathbb{T}) \to C(\mathbb{T})$ be defined by $\alpha(f)(z)=f(e^{-2\pi i\theta} z)$. I need to prove that $\alpha$ is an automorphism. It is clear that $\alpha$ is a surjective homomorphism. I want to prove the injectivity. If $u$ denotes the constant function $z$ on $\mathbb{T}$, the $\alpha(u)=e^{-2\pi i\theta}u$. Using this how can I conclude that $\alpha$ is injective?
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1Just show up the inverse of $\alpha$. – Berci Dec 18 '19 at 08:04
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Ya. The inverse is $\alpha^{-1}(f)(z)=f(e^{2 \pi i \theta}z)$. Am I right? Then what is the importance of $u$? – budi Dec 18 '19 at 08:06
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If $u$ is constant then $\alpha(u)=u$, not $e^{-i 2\pi \theta} , u$. – s.harp Dec 18 '19 at 09:33
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You don't have to use $u$ for injectivity. If $\alpha (f)=\alpha (g)$ then $f(e^{-2\pi i\theta} z)=g(e^{-2\pi i\theta} z)$ for all $z$. Changing $z$ to $e^{2\pi i\theta} z$ we get $f(z)=g(z)$ for all $z$ so $f=g$.
Kavi Rama Murthy
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