Continuity of the multiplicative inverse; $f:\Omega \rightarrow \mathbb{C}$, where $\Omega$ is a compact Hausdorff space.

Question

Is it true that if $f$ is continuous, then the multiplicative inverse $1/f$ (provided f is non-zero for any $x\in \Omega$) is also continuous? For some context, I am studying operator algebras, and without proof, my reference book said that for any $f\in C(\Omega),$ the spectrum $\sigma(f)$ is equal the image $f(\Omega).$ I think this is pretty easy to prove, if I can show that the multiplicative inverse of a continuous function is also continuous. Also, is the hypothesis that $\Omega$ is compact-Hausforff really necessary?

Answer 1

Note that $1/f = \xi f$ with $\xi : z \in \mathbb{C}^\times \mapsto \frac{1}{z} \in \mathbb{C}$. Since $\xi$ is continuous, so is $1/f$, regardless of the additional conditions on $\Omega$.

Edit: strictly speaking, one should say that $1/f = \xi \tilde{f}$ with $\tilde{f}$ the correstriction of $f$ to $\mathbb{C}^\times$ which exists (and is continuous) by hypothesis.