The characteristic polynomial map and its section

Idea. Continuity follows from the polynomial dependence. A right inverse is a continuous choice of a matrix with prescribed characteristic polynomial; the companion matrix does it.

Step 1: continuity. $\det(tI-A)$ is a polynomial in $t$ whose coefficients are polynomial in the entries $a_{ij}$ (expand the determinant by Leibniz). So each $c_k$ is a polynomial map, hence continuous (Problem 66). Explicitly $c_{n-1}=-\operatorname{tr}A$ and $c_0=(-1)^n\det A$.

Step 2: the companion matrix. For $c=(c_{n-1},\dots,c_0)\in\mathbb{R}^n$ let
$$C(c)=\begin{pmatrix}0&0&\cdots&0&-c_0\\1&0&\cdots&0&-c_1\\0&1&\cdots&0&-c_2\\\vdots&&\ddots&&\vdots\\0&0&\cdots&1&-c_{n-1}\end{pmatrix}.$$
The entries of $C(c)$ are $0$, $1$ or $-c_k$, so $c\mapsto C(c)$ is continuous (it is even affine).

Step 3: $\chi(C(c))=c$. Expanding $\det(tI-C(c))$ along the last column gives $t^n+c_{n-1}t^{n-1}+\dots+c_0$. (Check $n=2$: $C=\begin{pmatrix}0&-c_0\\1&-c_1\end{pmatrix}$, $\det(tI-C)=t(t+c_1)+c_0=t^2+c_1t+c_0$.)

Step 4: conclusion. $\chi\circ C=\mathrm{id}_{\mathbb{R}^n}$. So $\chi$ is surjective, and $C$ is a continuous right inverse (a continuous *section*).

Remark. There is no *linear* right inverse for $n\ge2$ in general, but there is a continuous one. Also, a continuous *left* inverse is impossible, because $\chi$ is far from injective (all conjugates of $A$ share the same characteristic polynomial). This map identifies the quotient "$M_n$ modulo conjugation" with the space of polynomials, at least on the open set of matrices with distinct eigenvalues.