Is $f:M_n(\mathbb{C})\longrightarrow M_n(\mathbb{C})$ continuous? I want to know whether this is absurd question or reasonable to ask: Let 
$f:M_n(\mathbb{C})\to M_n(\mathbb{C})$ be given by $f(A)= B$, where $B$ is a diagonal matrix having the same eigenvalues as $A$. Is $f$ continuous?
 A: There is no way to order the eigenvalues so that the function $f$ (as a function into $M_n({\mathbb C})$) is continuous.  Consider the matrices $A(t) = \pmatrix{0 & 1\cr e^{2it} & 0\cr}$.  Note that $A(0) = A(\pi)$.  The eigenvalues are $\pm e^{it}$.  But if you take the eigenvalue that is $1$ at $t=0$ and follow it continuously as $t$ goes from $0$ to $\pi$, it will be $-1$ at $t=\pi$.
A: I think what's behind the question is this:
Write 
$$
(X-z_1)\cdots(X-z_n)=X^n+c_1X^{n-1}+\cdots+c_n,\qquad(*)
$$
where $X$ is an indeterminate and
$$
z=(z_1,\dots,z_n),\quad c=(c_1,\dots,c_n)\in\mathbb C^n.
$$
For any $z$ there is a unique $c$ satisfying $(*)$. 
Moreover, the map  $z\mapsto c$, $\mathbb C^n\to\mathbb C^n$, is polynomial, and induces a continuous map $S_n\backslash\mathbb C^n\to\mathbb C^n$, where $S_n\backslash\mathbb C^n$ is the space of orbits of the symmetric group $S_n$ acting on $\mathbb C^n$ by permuting the coordinates. 
Conversely, let $c$ be given. By the Fundamental Theorem of Algebras, there is a $z$ satisfying $(*)$. 
Moreover, the $S_n$-orbit $S_nz$ of $z$ is depends only on $c$, and by Rouché's Theorem, the map $c\mapsto S_nz$, $\mathbb C^n\to S_n\backslash\mathbb C^n$ is continuous. 
Clearly the maps $S_nz\mapsto c$ and $c\mapsto S_nz$ are inverse. 
In conclusion, we have a homeomorphism 
$$
S_n\backslash\mathbb C^n\simeq\mathbb C^n.
$$
EDIT. What Robert Israel's answer shows is that there is no continuous section to the canonical projection $\mathbb C^n\to S_n\backslash\mathbb C^n$.
