# Continuous dependence of the roots of a polynomial on its coefficients

In their article "The roots of a polynomial vary continuously as a function of the coefficients" Gary Harris and Clyde Martin give a topological proof of the well-known theorem that the roots of a polynomial depend continuously on its coefficients. I don't understand the key moment of the proof of the "Treorem A" (stating that $\hat\sigma\colon\mathbb C^n/\sim\to\mathbb C^n$ is a homeomorphism).

First, the authors define the metric $d$ on $\mathbb C^n/\sim$ and prove that the topology induced by $d$ is the quotient topology with respect to $\pi\colon\mathbb C^n\to\mathbb C^n/\sim$. Then they show that $\hat\sigma|_{\overline{B(0,M)}}$ is a homeomorphism where $B(0,M)$ is the open ball about $0$ in $\mathbb C^n/\sim$ of radius $M$. Finally, they show that the map $\hat\sigma\colon\mathbb C^n/\sim\to\mathbb C^n$ is open. For this purpose the authors take an arbitrary open set $U\subset\mathbb C^n/\sim$, a point $x\in U$ and the open ball $B(x,\varepsilon)\subset U$ such that $\overline{B(x,\varepsilon)}\subset B(0,M)$ for some suitable values $\varepsilon$ and $M$. Then they write "Since $\hat\sigma$ is open on $B(0,M)$ it follows that $\hat\sigma(x)$ is the interior of $\hat\sigma(B(x,\varepsilon))$".

The last sentence I don't understand. By the above argument we know that $\hat\sigma|_{\overline{B(0,M)}}$ is a homeomorphis onto its image. Thus the image of any open set in $\overline{B(0,M)}$ is open in $\hat\sigma(\overline{B(0,M)})$. In particular the image of any open set in $B(0,M)$ is open in $\hat\sigma(B(0,M))$ but not necessarily in $\mathbb C^n$. It means that $\hat\sigma(B(x,\varepsilon))$ is a neighborhood of $x$ in $\hat\sigma(B(0,M))$. But why the set $\hat\sigma(B(x,\varepsilon))$ is open in $\mathbb C^n$? Why the authors conclude that the set $\hat\sigma(U)$ is open in $\mathbb C^n$? Thanks for helping.